148455
A thermodynamic process in which the system is insulated from the surroundings and no heat flows between the system and the surroundings is an
1 isothermal process
2 adiabatic process
3 isochoric process
4 isobaric process
Explanation:
B The thermodynamic process in which there is no exchange of heat from the system to its surrounding neither during expansion nor during compression is called adiabatic process. In an adiabatic process $\Delta Q=0$ - Isothermal process is a type of thermodynamic process in which the temperature $\mathrm{T}$ of a system remains constant $\Delta \mathrm{T}=0$ - A thermodynamic process in which the volume of the system is kept constant is called isochoric Process. $\Delta \mathrm{V}=0$ - Isobaric process is a type of thermodynamic process in which the pressure of the system remains constant. $\Delta \mathrm{P}=0$
J and K CET- 2011
Thermodynamics
148457
A certain mass of gas at NTP is expanded to three times its volume under adiabatic conditions. The resulting temperature of gas will be ( $\gamma$ for gas is 1.40)
1 $273 \times\left(\frac{1}{3}\right)^{1.4}$
2 $273 \times(3)^{0.4}$
3 $273 \times\left(\frac{1}{3}\right)^{0.4}$
4 $273 \times(3)^{1.4}$
Explanation:
C As we know that, according to adiabatic law $\mathrm{TV}^{\gamma-1}=$ Constant $\mathrm{T}_{1} \mathrm{~V}_{1}^{\gamma-1}=\mathrm{T}_{2} \mathrm{~V}_{2}^{\gamma-1}$ Given that $\gamma=1.40$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma-1}$ or, $\mathrm{T}_{2}=\mathrm{T}_{1}\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma-1}$ $\mathrm{~T}_{2}=273\left(\frac{\mathrm{V}_{1}}{3}\right)^{1.4-1}$ $\therefore \mathrm{T}_{2}=273\left(\frac{1}{3}\right)^{0.4}$
J and K CET- 2003
Thermodynamics
148460
In an adiabatic change, the pressure $P$ and temperature $T$ of a monoatomic gas are related by the relation $P \propto T^{c}$, where $c$ is equal to
1 $\frac{5}{3}$
2 $\frac{2}{5}$
3 $\frac{3}{5}$
4 $\frac{5}{2}$
Explanation:
D We know that, For adiabatic change - $\frac{\mathrm{T}^{\gamma}}{\mathrm{P}^{\gamma-1}}=\text { constant }$ $\mathrm{T}^{\gamma} \propto \mathrm{P}^{\gamma-1}$ $\mathrm{~T}^{\frac{\gamma}{\gamma-1}} \propto \mathrm{P}$ $\mathrm{P} \propto \mathrm{T}^{\frac{\gamma}{\gamma-1}}$ $\mathrm{c}=\frac{\gamma}{\gamma-1}=\frac{5 / 3}{5 / 3-1}$ $\mathrm{c}=\frac{5}{2}$
UP CPMT-2010
Thermodynamics
148462
Consider $n$ moles of an ideal gas. The gas expands adiabatically from initial temperature $T_{i}$ to final lower temperature $T_{f}$. The work done by gas is proportional to
1 Specific heat at constant pressure and difference between initial and final temperature
2 Specific heat at constant volume and difference between initial and final temperature
3 Only specific heat
4 Only difference between initial and final temperature
Explanation:
B Given, Initial temperature $=\mathrm{T}_{\mathrm{i}}$ final temperature $=\mathrm{T}_{f}$ as we know work done in an adiabatic expansion is given by - $\mathrm{W}=\frac{\mathrm{R}}{\gamma-1}\left[\mathrm{~T}_{\mathrm{i}}-\mathrm{T}_{f}\right]$ Where, $\mathrm{R}=\text { universal gas constant }$ $\gamma=\text { adiabatic constant }$ Now, universal gas constant is given by $\mathrm{R}=\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}$ Hence after putting the value of $\mathrm{R}$ in equation (i) we get, $\mathrm{W}=\frac{\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}}{\gamma-1}\left[\mathrm{~T}_{\mathrm{i}}-\mathrm{T}_{f}\right]$ $\because \quad \gamma=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}$, so we have $\mathrm{W}=\frac{\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}}{\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}-1}\left[\mathrm{~T}_{\mathrm{i}}-\mathrm{T}_{f}\right]$ $\mathrm{W}=\frac{\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}}{\left(\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}\right) / \mathrm{C}_{\mathrm{v}}}\left[\mathrm{T}_{\mathrm{i}}-\mathrm{T}_{f}\right]$ $\mathrm{W}=\mathrm{C}_{\mathrm{v}}\left(\mathrm{T}_{\mathrm{i}}-\mathrm{T}_{f}\right)$ Now from this result we can say that the work done by gas is proportional to specific heat at constant volume and difference between initial and final temperature.
TS EAMCET 31.07.2022
Thermodynamics
148465
An ideal monoatomic gas of volume $V$ is adiabatically expanded to a volume $3 \mathrm{~V}$ at $27^{\circ} \mathrm{C}$. The final temperature in Kelvins is: $\text { (use } \frac{C_{P}}{C_{V}}=5 / 3 \text { ) }$
1 144.2
2 170.3
3 50.4
4 100.2
Explanation:
A In an adiabatic process, $\mathrm{PV}^{\gamma}=\text { constant }$ $\mathrm{TV}^{\gamma-1}=\text { constant }$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma-1}$ For Monoatomic'gas $\gamma=\frac{5}{3}$ $\mathrm{~T}_{1}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$ $\mathrm{~T}_{2}=300\left(\frac{\mathrm{V}}{3 \mathrm{~V}}\right)^{\frac{5}{3}-1}=300\left(\frac{\mathrm{V}}{3 \mathrm{~V}}\right)^{2 / 3}$ $=300\left[\frac{1}{3}\right]^{2 / 3}$ $\mathrm{~T}_{2}{ }^{3}=(300)^{3}\left[\frac{1}{3}\right]^{\frac{2}{3} \times 3}$ $\mathrm{~T}_{2}{ }^{3}=27000000\left[\frac{1}{9}\right]$ $\mathrm{T}_{2}{ }^{3}=3000000$ $\mathrm{~T}_{2}=144.2 \mathrm{~K}$
148455
A thermodynamic process in which the system is insulated from the surroundings and no heat flows between the system and the surroundings is an
1 isothermal process
2 adiabatic process
3 isochoric process
4 isobaric process
Explanation:
B The thermodynamic process in which there is no exchange of heat from the system to its surrounding neither during expansion nor during compression is called adiabatic process. In an adiabatic process $\Delta Q=0$ - Isothermal process is a type of thermodynamic process in which the temperature $\mathrm{T}$ of a system remains constant $\Delta \mathrm{T}=0$ - A thermodynamic process in which the volume of the system is kept constant is called isochoric Process. $\Delta \mathrm{V}=0$ - Isobaric process is a type of thermodynamic process in which the pressure of the system remains constant. $\Delta \mathrm{P}=0$
J and K CET- 2011
Thermodynamics
148457
A certain mass of gas at NTP is expanded to three times its volume under adiabatic conditions. The resulting temperature of gas will be ( $\gamma$ for gas is 1.40)
1 $273 \times\left(\frac{1}{3}\right)^{1.4}$
2 $273 \times(3)^{0.4}$
3 $273 \times\left(\frac{1}{3}\right)^{0.4}$
4 $273 \times(3)^{1.4}$
Explanation:
C As we know that, according to adiabatic law $\mathrm{TV}^{\gamma-1}=$ Constant $\mathrm{T}_{1} \mathrm{~V}_{1}^{\gamma-1}=\mathrm{T}_{2} \mathrm{~V}_{2}^{\gamma-1}$ Given that $\gamma=1.40$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma-1}$ or, $\mathrm{T}_{2}=\mathrm{T}_{1}\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma-1}$ $\mathrm{~T}_{2}=273\left(\frac{\mathrm{V}_{1}}{3}\right)^{1.4-1}$ $\therefore \mathrm{T}_{2}=273\left(\frac{1}{3}\right)^{0.4}$
J and K CET- 2003
Thermodynamics
148460
In an adiabatic change, the pressure $P$ and temperature $T$ of a monoatomic gas are related by the relation $P \propto T^{c}$, where $c$ is equal to
1 $\frac{5}{3}$
2 $\frac{2}{5}$
3 $\frac{3}{5}$
4 $\frac{5}{2}$
Explanation:
D We know that, For adiabatic change - $\frac{\mathrm{T}^{\gamma}}{\mathrm{P}^{\gamma-1}}=\text { constant }$ $\mathrm{T}^{\gamma} \propto \mathrm{P}^{\gamma-1}$ $\mathrm{~T}^{\frac{\gamma}{\gamma-1}} \propto \mathrm{P}$ $\mathrm{P} \propto \mathrm{T}^{\frac{\gamma}{\gamma-1}}$ $\mathrm{c}=\frac{\gamma}{\gamma-1}=\frac{5 / 3}{5 / 3-1}$ $\mathrm{c}=\frac{5}{2}$
UP CPMT-2010
Thermodynamics
148462
Consider $n$ moles of an ideal gas. The gas expands adiabatically from initial temperature $T_{i}$ to final lower temperature $T_{f}$. The work done by gas is proportional to
1 Specific heat at constant pressure and difference between initial and final temperature
2 Specific heat at constant volume and difference between initial and final temperature
3 Only specific heat
4 Only difference between initial and final temperature
Explanation:
B Given, Initial temperature $=\mathrm{T}_{\mathrm{i}}$ final temperature $=\mathrm{T}_{f}$ as we know work done in an adiabatic expansion is given by - $\mathrm{W}=\frac{\mathrm{R}}{\gamma-1}\left[\mathrm{~T}_{\mathrm{i}}-\mathrm{T}_{f}\right]$ Where, $\mathrm{R}=\text { universal gas constant }$ $\gamma=\text { adiabatic constant }$ Now, universal gas constant is given by $\mathrm{R}=\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}$ Hence after putting the value of $\mathrm{R}$ in equation (i) we get, $\mathrm{W}=\frac{\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}}{\gamma-1}\left[\mathrm{~T}_{\mathrm{i}}-\mathrm{T}_{f}\right]$ $\because \quad \gamma=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}$, so we have $\mathrm{W}=\frac{\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}}{\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}-1}\left[\mathrm{~T}_{\mathrm{i}}-\mathrm{T}_{f}\right]$ $\mathrm{W}=\frac{\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}}{\left(\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}\right) / \mathrm{C}_{\mathrm{v}}}\left[\mathrm{T}_{\mathrm{i}}-\mathrm{T}_{f}\right]$ $\mathrm{W}=\mathrm{C}_{\mathrm{v}}\left(\mathrm{T}_{\mathrm{i}}-\mathrm{T}_{f}\right)$ Now from this result we can say that the work done by gas is proportional to specific heat at constant volume and difference between initial and final temperature.
TS EAMCET 31.07.2022
Thermodynamics
148465
An ideal monoatomic gas of volume $V$ is adiabatically expanded to a volume $3 \mathrm{~V}$ at $27^{\circ} \mathrm{C}$. The final temperature in Kelvins is: $\text { (use } \frac{C_{P}}{C_{V}}=5 / 3 \text { ) }$
1 144.2
2 170.3
3 50.4
4 100.2
Explanation:
A In an adiabatic process, $\mathrm{PV}^{\gamma}=\text { constant }$ $\mathrm{TV}^{\gamma-1}=\text { constant }$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma-1}$ For Monoatomic'gas $\gamma=\frac{5}{3}$ $\mathrm{~T}_{1}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$ $\mathrm{~T}_{2}=300\left(\frac{\mathrm{V}}{3 \mathrm{~V}}\right)^{\frac{5}{3}-1}=300\left(\frac{\mathrm{V}}{3 \mathrm{~V}}\right)^{2 / 3}$ $=300\left[\frac{1}{3}\right]^{2 / 3}$ $\mathrm{~T}_{2}{ }^{3}=(300)^{3}\left[\frac{1}{3}\right]^{\frac{2}{3} \times 3}$ $\mathrm{~T}_{2}{ }^{3}=27000000\left[\frac{1}{9}\right]$ $\mathrm{T}_{2}{ }^{3}=3000000$ $\mathrm{~T}_{2}=144.2 \mathrm{~K}$
148455
A thermodynamic process in which the system is insulated from the surroundings and no heat flows between the system and the surroundings is an
1 isothermal process
2 adiabatic process
3 isochoric process
4 isobaric process
Explanation:
B The thermodynamic process in which there is no exchange of heat from the system to its surrounding neither during expansion nor during compression is called adiabatic process. In an adiabatic process $\Delta Q=0$ - Isothermal process is a type of thermodynamic process in which the temperature $\mathrm{T}$ of a system remains constant $\Delta \mathrm{T}=0$ - A thermodynamic process in which the volume of the system is kept constant is called isochoric Process. $\Delta \mathrm{V}=0$ - Isobaric process is a type of thermodynamic process in which the pressure of the system remains constant. $\Delta \mathrm{P}=0$
J and K CET- 2011
Thermodynamics
148457
A certain mass of gas at NTP is expanded to three times its volume under adiabatic conditions. The resulting temperature of gas will be ( $\gamma$ for gas is 1.40)
1 $273 \times\left(\frac{1}{3}\right)^{1.4}$
2 $273 \times(3)^{0.4}$
3 $273 \times\left(\frac{1}{3}\right)^{0.4}$
4 $273 \times(3)^{1.4}$
Explanation:
C As we know that, according to adiabatic law $\mathrm{TV}^{\gamma-1}=$ Constant $\mathrm{T}_{1} \mathrm{~V}_{1}^{\gamma-1}=\mathrm{T}_{2} \mathrm{~V}_{2}^{\gamma-1}$ Given that $\gamma=1.40$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma-1}$ or, $\mathrm{T}_{2}=\mathrm{T}_{1}\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma-1}$ $\mathrm{~T}_{2}=273\left(\frac{\mathrm{V}_{1}}{3}\right)^{1.4-1}$ $\therefore \mathrm{T}_{2}=273\left(\frac{1}{3}\right)^{0.4}$
J and K CET- 2003
Thermodynamics
148460
In an adiabatic change, the pressure $P$ and temperature $T$ of a monoatomic gas are related by the relation $P \propto T^{c}$, where $c$ is equal to
1 $\frac{5}{3}$
2 $\frac{2}{5}$
3 $\frac{3}{5}$
4 $\frac{5}{2}$
Explanation:
D We know that, For adiabatic change - $\frac{\mathrm{T}^{\gamma}}{\mathrm{P}^{\gamma-1}}=\text { constant }$ $\mathrm{T}^{\gamma} \propto \mathrm{P}^{\gamma-1}$ $\mathrm{~T}^{\frac{\gamma}{\gamma-1}} \propto \mathrm{P}$ $\mathrm{P} \propto \mathrm{T}^{\frac{\gamma}{\gamma-1}}$ $\mathrm{c}=\frac{\gamma}{\gamma-1}=\frac{5 / 3}{5 / 3-1}$ $\mathrm{c}=\frac{5}{2}$
UP CPMT-2010
Thermodynamics
148462
Consider $n$ moles of an ideal gas. The gas expands adiabatically from initial temperature $T_{i}$ to final lower temperature $T_{f}$. The work done by gas is proportional to
1 Specific heat at constant pressure and difference between initial and final temperature
2 Specific heat at constant volume and difference between initial and final temperature
3 Only specific heat
4 Only difference between initial and final temperature
Explanation:
B Given, Initial temperature $=\mathrm{T}_{\mathrm{i}}$ final temperature $=\mathrm{T}_{f}$ as we know work done in an adiabatic expansion is given by - $\mathrm{W}=\frac{\mathrm{R}}{\gamma-1}\left[\mathrm{~T}_{\mathrm{i}}-\mathrm{T}_{f}\right]$ Where, $\mathrm{R}=\text { universal gas constant }$ $\gamma=\text { adiabatic constant }$ Now, universal gas constant is given by $\mathrm{R}=\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}$ Hence after putting the value of $\mathrm{R}$ in equation (i) we get, $\mathrm{W}=\frac{\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}}{\gamma-1}\left[\mathrm{~T}_{\mathrm{i}}-\mathrm{T}_{f}\right]$ $\because \quad \gamma=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}$, so we have $\mathrm{W}=\frac{\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}}{\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}-1}\left[\mathrm{~T}_{\mathrm{i}}-\mathrm{T}_{f}\right]$ $\mathrm{W}=\frac{\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}}{\left(\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}\right) / \mathrm{C}_{\mathrm{v}}}\left[\mathrm{T}_{\mathrm{i}}-\mathrm{T}_{f}\right]$ $\mathrm{W}=\mathrm{C}_{\mathrm{v}}\left(\mathrm{T}_{\mathrm{i}}-\mathrm{T}_{f}\right)$ Now from this result we can say that the work done by gas is proportional to specific heat at constant volume and difference between initial and final temperature.
TS EAMCET 31.07.2022
Thermodynamics
148465
An ideal monoatomic gas of volume $V$ is adiabatically expanded to a volume $3 \mathrm{~V}$ at $27^{\circ} \mathrm{C}$. The final temperature in Kelvins is: $\text { (use } \frac{C_{P}}{C_{V}}=5 / 3 \text { ) }$
1 144.2
2 170.3
3 50.4
4 100.2
Explanation:
A In an adiabatic process, $\mathrm{PV}^{\gamma}=\text { constant }$ $\mathrm{TV}^{\gamma-1}=\text { constant }$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma-1}$ For Monoatomic'gas $\gamma=\frac{5}{3}$ $\mathrm{~T}_{1}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$ $\mathrm{~T}_{2}=300\left(\frac{\mathrm{V}}{3 \mathrm{~V}}\right)^{\frac{5}{3}-1}=300\left(\frac{\mathrm{V}}{3 \mathrm{~V}}\right)^{2 / 3}$ $=300\left[\frac{1}{3}\right]^{2 / 3}$ $\mathrm{~T}_{2}{ }^{3}=(300)^{3}\left[\frac{1}{3}\right]^{\frac{2}{3} \times 3}$ $\mathrm{~T}_{2}{ }^{3}=27000000\left[\frac{1}{9}\right]$ $\mathrm{T}_{2}{ }^{3}=3000000$ $\mathrm{~T}_{2}=144.2 \mathrm{~K}$
148455
A thermodynamic process in which the system is insulated from the surroundings and no heat flows between the system and the surroundings is an
1 isothermal process
2 adiabatic process
3 isochoric process
4 isobaric process
Explanation:
B The thermodynamic process in which there is no exchange of heat from the system to its surrounding neither during expansion nor during compression is called adiabatic process. In an adiabatic process $\Delta Q=0$ - Isothermal process is a type of thermodynamic process in which the temperature $\mathrm{T}$ of a system remains constant $\Delta \mathrm{T}=0$ - A thermodynamic process in which the volume of the system is kept constant is called isochoric Process. $\Delta \mathrm{V}=0$ - Isobaric process is a type of thermodynamic process in which the pressure of the system remains constant. $\Delta \mathrm{P}=0$
J and K CET- 2011
Thermodynamics
148457
A certain mass of gas at NTP is expanded to three times its volume under adiabatic conditions. The resulting temperature of gas will be ( $\gamma$ for gas is 1.40)
1 $273 \times\left(\frac{1}{3}\right)^{1.4}$
2 $273 \times(3)^{0.4}$
3 $273 \times\left(\frac{1}{3}\right)^{0.4}$
4 $273 \times(3)^{1.4}$
Explanation:
C As we know that, according to adiabatic law $\mathrm{TV}^{\gamma-1}=$ Constant $\mathrm{T}_{1} \mathrm{~V}_{1}^{\gamma-1}=\mathrm{T}_{2} \mathrm{~V}_{2}^{\gamma-1}$ Given that $\gamma=1.40$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma-1}$ or, $\mathrm{T}_{2}=\mathrm{T}_{1}\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma-1}$ $\mathrm{~T}_{2}=273\left(\frac{\mathrm{V}_{1}}{3}\right)^{1.4-1}$ $\therefore \mathrm{T}_{2}=273\left(\frac{1}{3}\right)^{0.4}$
J and K CET- 2003
Thermodynamics
148460
In an adiabatic change, the pressure $P$ and temperature $T$ of a monoatomic gas are related by the relation $P \propto T^{c}$, where $c$ is equal to
1 $\frac{5}{3}$
2 $\frac{2}{5}$
3 $\frac{3}{5}$
4 $\frac{5}{2}$
Explanation:
D We know that, For adiabatic change - $\frac{\mathrm{T}^{\gamma}}{\mathrm{P}^{\gamma-1}}=\text { constant }$ $\mathrm{T}^{\gamma} \propto \mathrm{P}^{\gamma-1}$ $\mathrm{~T}^{\frac{\gamma}{\gamma-1}} \propto \mathrm{P}$ $\mathrm{P} \propto \mathrm{T}^{\frac{\gamma}{\gamma-1}}$ $\mathrm{c}=\frac{\gamma}{\gamma-1}=\frac{5 / 3}{5 / 3-1}$ $\mathrm{c}=\frac{5}{2}$
UP CPMT-2010
Thermodynamics
148462
Consider $n$ moles of an ideal gas. The gas expands adiabatically from initial temperature $T_{i}$ to final lower temperature $T_{f}$. The work done by gas is proportional to
1 Specific heat at constant pressure and difference between initial and final temperature
2 Specific heat at constant volume and difference between initial and final temperature
3 Only specific heat
4 Only difference between initial and final temperature
Explanation:
B Given, Initial temperature $=\mathrm{T}_{\mathrm{i}}$ final temperature $=\mathrm{T}_{f}$ as we know work done in an adiabatic expansion is given by - $\mathrm{W}=\frac{\mathrm{R}}{\gamma-1}\left[\mathrm{~T}_{\mathrm{i}}-\mathrm{T}_{f}\right]$ Where, $\mathrm{R}=\text { universal gas constant }$ $\gamma=\text { adiabatic constant }$ Now, universal gas constant is given by $\mathrm{R}=\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}$ Hence after putting the value of $\mathrm{R}$ in equation (i) we get, $\mathrm{W}=\frac{\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}}{\gamma-1}\left[\mathrm{~T}_{\mathrm{i}}-\mathrm{T}_{f}\right]$ $\because \quad \gamma=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}$, so we have $\mathrm{W}=\frac{\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}}{\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}-1}\left[\mathrm{~T}_{\mathrm{i}}-\mathrm{T}_{f}\right]$ $\mathrm{W}=\frac{\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}}{\left(\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}\right) / \mathrm{C}_{\mathrm{v}}}\left[\mathrm{T}_{\mathrm{i}}-\mathrm{T}_{f}\right]$ $\mathrm{W}=\mathrm{C}_{\mathrm{v}}\left(\mathrm{T}_{\mathrm{i}}-\mathrm{T}_{f}\right)$ Now from this result we can say that the work done by gas is proportional to specific heat at constant volume and difference between initial and final temperature.
TS EAMCET 31.07.2022
Thermodynamics
148465
An ideal monoatomic gas of volume $V$ is adiabatically expanded to a volume $3 \mathrm{~V}$ at $27^{\circ} \mathrm{C}$. The final temperature in Kelvins is: $\text { (use } \frac{C_{P}}{C_{V}}=5 / 3 \text { ) }$
1 144.2
2 170.3
3 50.4
4 100.2
Explanation:
A In an adiabatic process, $\mathrm{PV}^{\gamma}=\text { constant }$ $\mathrm{TV}^{\gamma-1}=\text { constant }$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma-1}$ For Monoatomic'gas $\gamma=\frac{5}{3}$ $\mathrm{~T}_{1}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$ $\mathrm{~T}_{2}=300\left(\frac{\mathrm{V}}{3 \mathrm{~V}}\right)^{\frac{5}{3}-1}=300\left(\frac{\mathrm{V}}{3 \mathrm{~V}}\right)^{2 / 3}$ $=300\left[\frac{1}{3}\right]^{2 / 3}$ $\mathrm{~T}_{2}{ }^{3}=(300)^{3}\left[\frac{1}{3}\right]^{\frac{2}{3} \times 3}$ $\mathrm{~T}_{2}{ }^{3}=27000000\left[\frac{1}{9}\right]$ $\mathrm{T}_{2}{ }^{3}=3000000$ $\mathrm{~T}_{2}=144.2 \mathrm{~K}$
148455
A thermodynamic process in which the system is insulated from the surroundings and no heat flows between the system and the surroundings is an
1 isothermal process
2 adiabatic process
3 isochoric process
4 isobaric process
Explanation:
B The thermodynamic process in which there is no exchange of heat from the system to its surrounding neither during expansion nor during compression is called adiabatic process. In an adiabatic process $\Delta Q=0$ - Isothermal process is a type of thermodynamic process in which the temperature $\mathrm{T}$ of a system remains constant $\Delta \mathrm{T}=0$ - A thermodynamic process in which the volume of the system is kept constant is called isochoric Process. $\Delta \mathrm{V}=0$ - Isobaric process is a type of thermodynamic process in which the pressure of the system remains constant. $\Delta \mathrm{P}=0$
J and K CET- 2011
Thermodynamics
148457
A certain mass of gas at NTP is expanded to three times its volume under adiabatic conditions. The resulting temperature of gas will be ( $\gamma$ for gas is 1.40)
1 $273 \times\left(\frac{1}{3}\right)^{1.4}$
2 $273 \times(3)^{0.4}$
3 $273 \times\left(\frac{1}{3}\right)^{0.4}$
4 $273 \times(3)^{1.4}$
Explanation:
C As we know that, according to adiabatic law $\mathrm{TV}^{\gamma-1}=$ Constant $\mathrm{T}_{1} \mathrm{~V}_{1}^{\gamma-1}=\mathrm{T}_{2} \mathrm{~V}_{2}^{\gamma-1}$ Given that $\gamma=1.40$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma-1}$ or, $\mathrm{T}_{2}=\mathrm{T}_{1}\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma-1}$ $\mathrm{~T}_{2}=273\left(\frac{\mathrm{V}_{1}}{3}\right)^{1.4-1}$ $\therefore \mathrm{T}_{2}=273\left(\frac{1}{3}\right)^{0.4}$
J and K CET- 2003
Thermodynamics
148460
In an adiabatic change, the pressure $P$ and temperature $T$ of a monoatomic gas are related by the relation $P \propto T^{c}$, where $c$ is equal to
1 $\frac{5}{3}$
2 $\frac{2}{5}$
3 $\frac{3}{5}$
4 $\frac{5}{2}$
Explanation:
D We know that, For adiabatic change - $\frac{\mathrm{T}^{\gamma}}{\mathrm{P}^{\gamma-1}}=\text { constant }$ $\mathrm{T}^{\gamma} \propto \mathrm{P}^{\gamma-1}$ $\mathrm{~T}^{\frac{\gamma}{\gamma-1}} \propto \mathrm{P}$ $\mathrm{P} \propto \mathrm{T}^{\frac{\gamma}{\gamma-1}}$ $\mathrm{c}=\frac{\gamma}{\gamma-1}=\frac{5 / 3}{5 / 3-1}$ $\mathrm{c}=\frac{5}{2}$
UP CPMT-2010
Thermodynamics
148462
Consider $n$ moles of an ideal gas. The gas expands adiabatically from initial temperature $T_{i}$ to final lower temperature $T_{f}$. The work done by gas is proportional to
1 Specific heat at constant pressure and difference between initial and final temperature
2 Specific heat at constant volume and difference between initial and final temperature
3 Only specific heat
4 Only difference between initial and final temperature
Explanation:
B Given, Initial temperature $=\mathrm{T}_{\mathrm{i}}$ final temperature $=\mathrm{T}_{f}$ as we know work done in an adiabatic expansion is given by - $\mathrm{W}=\frac{\mathrm{R}}{\gamma-1}\left[\mathrm{~T}_{\mathrm{i}}-\mathrm{T}_{f}\right]$ Where, $\mathrm{R}=\text { universal gas constant }$ $\gamma=\text { adiabatic constant }$ Now, universal gas constant is given by $\mathrm{R}=\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}$ Hence after putting the value of $\mathrm{R}$ in equation (i) we get, $\mathrm{W}=\frac{\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}}{\gamma-1}\left[\mathrm{~T}_{\mathrm{i}}-\mathrm{T}_{f}\right]$ $\because \quad \gamma=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}$, so we have $\mathrm{W}=\frac{\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}}{\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}-1}\left[\mathrm{~T}_{\mathrm{i}}-\mathrm{T}_{f}\right]$ $\mathrm{W}=\frac{\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}}{\left(\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}\right) / \mathrm{C}_{\mathrm{v}}}\left[\mathrm{T}_{\mathrm{i}}-\mathrm{T}_{f}\right]$ $\mathrm{W}=\mathrm{C}_{\mathrm{v}}\left(\mathrm{T}_{\mathrm{i}}-\mathrm{T}_{f}\right)$ Now from this result we can say that the work done by gas is proportional to specific heat at constant volume and difference between initial and final temperature.
TS EAMCET 31.07.2022
Thermodynamics
148465
An ideal monoatomic gas of volume $V$ is adiabatically expanded to a volume $3 \mathrm{~V}$ at $27^{\circ} \mathrm{C}$. The final temperature in Kelvins is: $\text { (use } \frac{C_{P}}{C_{V}}=5 / 3 \text { ) }$
1 144.2
2 170.3
3 50.4
4 100.2
Explanation:
A In an adiabatic process, $\mathrm{PV}^{\gamma}=\text { constant }$ $\mathrm{TV}^{\gamma-1}=\text { constant }$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma-1}$ For Monoatomic'gas $\gamma=\frac{5}{3}$ $\mathrm{~T}_{1}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$ $\mathrm{~T}_{2}=300\left(\frac{\mathrm{V}}{3 \mathrm{~V}}\right)^{\frac{5}{3}-1}=300\left(\frac{\mathrm{V}}{3 \mathrm{~V}}\right)^{2 / 3}$ $=300\left[\frac{1}{3}\right]^{2 / 3}$ $\mathrm{~T}_{2}{ }^{3}=(300)^{3}\left[\frac{1}{3}\right]^{\frac{2}{3} \times 3}$ $\mathrm{~T}_{2}{ }^{3}=27000000\left[\frac{1}{9}\right]$ $\mathrm{T}_{2}{ }^{3}=3000000$ $\mathrm{~T}_{2}=144.2 \mathrm{~K}$
