148692
The change in the entropy of a 1 mole of an ideal gas which went through an isothermal process from an initial state $\left(\mathrm{P}_{1}, \mathrm{~V}_{1}, \mathrm{~T}\right)$ to the final state $\left(P_{2}, V_{2}, T\right)$ is equal to :
D Work done by $\mathrm{n}$ moles of the gas volume change from $V_{1}$ to $V_{2}$, $\mathrm{W}=\mathrm{nRT} \ln \frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}} \quad(\because \mathrm{n}=1)$ $\mathrm{W}=\mathrm{RT} \ln \frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}$ First law of thermodynamics, $\Delta \mathrm{Q}=\Delta \mathrm{U}+\mathrm{W} \quad \text { (for isothermal process, } \Delta \mathrm{U}=0)$ $\Delta \mathrm{Q}=\mathrm{W}$ From second law of thermodynamics, $\Delta S=\frac{\Delta Q}{T}=\frac{R T \ln \frac{V_{2}}{V_{1}}}{T}=R \ln \frac{V_{2}}{V_{1}}$
BCECE-2006
Thermodynamics
148693
If ' $\Delta Q$ ' is the amount of heat supplied to ' $n$ ' moles of a diatomic gas at constant pressure, ' $\Delta \mathbf{U}$ ' is the change in internal energy and ' $\Delta \mathbf{W}$ ' is the work done, then $\Delta W: \Delta \mathrm{U}: \Delta \mathbf{Q}$ is
1 $1: 2: 3$
2 $2: 5: 7$
3 $2: 3: 4$
4 $5: 7: 9$
Explanation:
B We know, At constant pressure $\Delta \mathrm{Q}=\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{T}$ $\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}$ $\mathrm{W}=\mathrm{P} \Delta \mathrm{V}=\mathrm{nR} \Delta \mathrm{T}$ $\therefore \quad \Delta \mathrm{Q}: \Delta \mathrm{U}: \mathrm{W}=\mathrm{C}_{\mathrm{p}}: \mathrm{C}_{\mathrm{v}}: \mathrm{R}$ For diatomic gas, $\mathrm{f}=5$ Gas is ideal and diatomic in nature. $\therefore \quad \mathrm{C}_{\mathrm{v}}=\frac{\mathrm{fR}}{2}=\frac{5 \mathrm{R}}{2}$ $\because \quad \mathrm{C}_{\mathrm{p}}=\mathrm{C}_{\mathrm{v}}+\mathrm{R}=\frac{5 \mathrm{R}}{2}+\mathrm{R}$ $=\frac{7 \mathrm{R}}{2}$ $\therefore \quad \Delta \mathrm{Q}: \Delta \mathrm{U}: \mathrm{W}=7: 5: 2$ Hence, $\mathrm{W}: \Delta \mathrm{U}: \Delta \mathrm{Q}=2: 5: 7$
MHT-CET 2020
Thermodynamics
148696
Find the ratio of $\frac{\Delta Q}{\Delta W}$ in an isobaric process, the ratio of molar specific heat capacities of the gas used is $\frac{C_{p}}{C_{v}}=\gamma$
1 $\frac{\gamma}{\gamma-1}$
2 $\frac{\gamma-1}{\gamma}$
3 $\frac{\gamma+1}{\gamma}$
4 $\frac{\gamma}{\gamma+1}$
Explanation:
A We know that, $\Delta \mathrm{Q}=\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{T}$ and $\quad \Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}$ Taking ratio of equation (i) and (ii), we get- $\frac{\Delta \mathrm{Q}}{\Delta \mathrm{U}}=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}$ $\because \quad \frac{\Delta \mathrm{Q}}{\Delta \mathrm{U}}=\gamma$ $\frac{\Delta \mathrm{Q}}{\Delta \mathrm{W}}=\frac{\Delta \mathrm{Q}}{\Delta \mathrm{Q}-\Delta \mathrm{U}}=\frac{1}{1-\frac{\Delta \mathrm{U}}{\Delta \mathrm{Q}}}$ $=\frac{1}{1-\frac{1}{\gamma}}$ $\frac{\Delta \mathrm{Q}}{\Delta \mathrm{W}}=\frac{\gamma}{\gamma-1}$
AP EAMCET-05.10.2021
Thermodynamics
148697
No process is possible whose sole result is the transfer of heat from a colder object to a hotter object. This is Clausius statement for
1 Zeroth law of thermodynamics
2 First law of thermodynamics
3 Second law of thermodynamics
4 Carnot's theorem
5 Principle of refrigeration
Explanation:
C There are two statements of second law of thermodynamics which are (1) Kelvin-Plank statement (2) Clausius statement Kelvin-Plank statement: It is impossible for a heat engine to produce a network in a complete cycle if it exchange heat only with a single reservoir. Clausius's statement: It is impossible to design a device, which works on a cycle and produces no effect other than heat transfer from a cold body to hot body.
Kerala CEE 04.07.2022
Thermodynamics
148700
A liquid of mass $m$ and specific heat $s$ is heated to a temperature $T$. Another liquid of mass $\mathrm{m} / \mathbf{2}$ and specific heat $2 \mathrm{~s}$ is heated to temperature 2T. If these two liquids are mixed, the resultant temperature of the mixture will be
1 $\frac{2}{3} \mathrm{~T}$
2 $\frac{8}{5} \mathrm{~T}$
3 $\frac{3}{4} \mathrm{~T}$
4 $\frac{3}{2} \mathrm{~T}$
Explanation:
D Given, Mass of liquid $\left(\mathrm{m}_{1}\right)=\mathrm{m}$ Specific heat of liquid $\left(\mathrm{C}_{1}\right)=\mathrm{s}$ Initial temperature of liquid $\left(\mathrm{T}_{1}\right)=\mathrm{T}$ Mass of liquid $\left(\mathrm{m}_{2}\right)=\mathrm{m} / 2$ Specific heat of liquid $\left(\mathrm{C}_{2}\right)=2 \mathrm{~s}$ Initial temperature of liquid $\left(\mathrm{T}_{2}\right)=2 \mathrm{~T}$ Let the resultant temperature of the mixture $=T_{m}$ Heat lost by liquid $1=$ Heat gained by liquid 2 $\mathrm{m}_{1} \mathrm{C}_{1}\left(\mathrm{~T}_{1}-\mathrm{T}_{\mathrm{m}}\right)=\mathrm{m}_{2} \mathrm{C}_{2}\left(\mathrm{~T}_{\mathrm{m}}-\mathrm{T}_{2}\right)$ $\mathrm{m} \times \mathrm{s} \times\left(\mathrm{T}-\mathrm{T}_{\mathrm{m}}\right)=\frac{\mathrm{m}}{2} \times 2 \mathrm{~s} \times\left(\mathrm{T}_{\mathrm{m}}-2 \mathrm{~T}\right)$ $\mathrm{T}-\mathrm{T}_{\mathrm{m}}=\mathrm{T}_{\mathrm{m}}-2 \mathrm{~T}$ $\mathrm{T}+2 \mathrm{~T}=2 \mathrm{~T}_{\mathrm{m}}$ $3 \mathrm{~T}=2 \mathrm{~T}_{\mathrm{m}}$ $\mathrm{T}_{\mathrm{m}}=\frac{3}{2} \mathrm{~T}$
148692
The change in the entropy of a 1 mole of an ideal gas which went through an isothermal process from an initial state $\left(\mathrm{P}_{1}, \mathrm{~V}_{1}, \mathrm{~T}\right)$ to the final state $\left(P_{2}, V_{2}, T\right)$ is equal to :
D Work done by $\mathrm{n}$ moles of the gas volume change from $V_{1}$ to $V_{2}$, $\mathrm{W}=\mathrm{nRT} \ln \frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}} \quad(\because \mathrm{n}=1)$ $\mathrm{W}=\mathrm{RT} \ln \frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}$ First law of thermodynamics, $\Delta \mathrm{Q}=\Delta \mathrm{U}+\mathrm{W} \quad \text { (for isothermal process, } \Delta \mathrm{U}=0)$ $\Delta \mathrm{Q}=\mathrm{W}$ From second law of thermodynamics, $\Delta S=\frac{\Delta Q}{T}=\frac{R T \ln \frac{V_{2}}{V_{1}}}{T}=R \ln \frac{V_{2}}{V_{1}}$
BCECE-2006
Thermodynamics
148693
If ' $\Delta Q$ ' is the amount of heat supplied to ' $n$ ' moles of a diatomic gas at constant pressure, ' $\Delta \mathbf{U}$ ' is the change in internal energy and ' $\Delta \mathbf{W}$ ' is the work done, then $\Delta W: \Delta \mathrm{U}: \Delta \mathbf{Q}$ is
1 $1: 2: 3$
2 $2: 5: 7$
3 $2: 3: 4$
4 $5: 7: 9$
Explanation:
B We know, At constant pressure $\Delta \mathrm{Q}=\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{T}$ $\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}$ $\mathrm{W}=\mathrm{P} \Delta \mathrm{V}=\mathrm{nR} \Delta \mathrm{T}$ $\therefore \quad \Delta \mathrm{Q}: \Delta \mathrm{U}: \mathrm{W}=\mathrm{C}_{\mathrm{p}}: \mathrm{C}_{\mathrm{v}}: \mathrm{R}$ For diatomic gas, $\mathrm{f}=5$ Gas is ideal and diatomic in nature. $\therefore \quad \mathrm{C}_{\mathrm{v}}=\frac{\mathrm{fR}}{2}=\frac{5 \mathrm{R}}{2}$ $\because \quad \mathrm{C}_{\mathrm{p}}=\mathrm{C}_{\mathrm{v}}+\mathrm{R}=\frac{5 \mathrm{R}}{2}+\mathrm{R}$ $=\frac{7 \mathrm{R}}{2}$ $\therefore \quad \Delta \mathrm{Q}: \Delta \mathrm{U}: \mathrm{W}=7: 5: 2$ Hence, $\mathrm{W}: \Delta \mathrm{U}: \Delta \mathrm{Q}=2: 5: 7$
MHT-CET 2020
Thermodynamics
148696
Find the ratio of $\frac{\Delta Q}{\Delta W}$ in an isobaric process, the ratio of molar specific heat capacities of the gas used is $\frac{C_{p}}{C_{v}}=\gamma$
1 $\frac{\gamma}{\gamma-1}$
2 $\frac{\gamma-1}{\gamma}$
3 $\frac{\gamma+1}{\gamma}$
4 $\frac{\gamma}{\gamma+1}$
Explanation:
A We know that, $\Delta \mathrm{Q}=\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{T}$ and $\quad \Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}$ Taking ratio of equation (i) and (ii), we get- $\frac{\Delta \mathrm{Q}}{\Delta \mathrm{U}}=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}$ $\because \quad \frac{\Delta \mathrm{Q}}{\Delta \mathrm{U}}=\gamma$ $\frac{\Delta \mathrm{Q}}{\Delta \mathrm{W}}=\frac{\Delta \mathrm{Q}}{\Delta \mathrm{Q}-\Delta \mathrm{U}}=\frac{1}{1-\frac{\Delta \mathrm{U}}{\Delta \mathrm{Q}}}$ $=\frac{1}{1-\frac{1}{\gamma}}$ $\frac{\Delta \mathrm{Q}}{\Delta \mathrm{W}}=\frac{\gamma}{\gamma-1}$
AP EAMCET-05.10.2021
Thermodynamics
148697
No process is possible whose sole result is the transfer of heat from a colder object to a hotter object. This is Clausius statement for
1 Zeroth law of thermodynamics
2 First law of thermodynamics
3 Second law of thermodynamics
4 Carnot's theorem
5 Principle of refrigeration
Explanation:
C There are two statements of second law of thermodynamics which are (1) Kelvin-Plank statement (2) Clausius statement Kelvin-Plank statement: It is impossible for a heat engine to produce a network in a complete cycle if it exchange heat only with a single reservoir. Clausius's statement: It is impossible to design a device, which works on a cycle and produces no effect other than heat transfer from a cold body to hot body.
Kerala CEE 04.07.2022
Thermodynamics
148700
A liquid of mass $m$ and specific heat $s$ is heated to a temperature $T$. Another liquid of mass $\mathrm{m} / \mathbf{2}$ and specific heat $2 \mathrm{~s}$ is heated to temperature 2T. If these two liquids are mixed, the resultant temperature of the mixture will be
1 $\frac{2}{3} \mathrm{~T}$
2 $\frac{8}{5} \mathrm{~T}$
3 $\frac{3}{4} \mathrm{~T}$
4 $\frac{3}{2} \mathrm{~T}$
Explanation:
D Given, Mass of liquid $\left(\mathrm{m}_{1}\right)=\mathrm{m}$ Specific heat of liquid $\left(\mathrm{C}_{1}\right)=\mathrm{s}$ Initial temperature of liquid $\left(\mathrm{T}_{1}\right)=\mathrm{T}$ Mass of liquid $\left(\mathrm{m}_{2}\right)=\mathrm{m} / 2$ Specific heat of liquid $\left(\mathrm{C}_{2}\right)=2 \mathrm{~s}$ Initial temperature of liquid $\left(\mathrm{T}_{2}\right)=2 \mathrm{~T}$ Let the resultant temperature of the mixture $=T_{m}$ Heat lost by liquid $1=$ Heat gained by liquid 2 $\mathrm{m}_{1} \mathrm{C}_{1}\left(\mathrm{~T}_{1}-\mathrm{T}_{\mathrm{m}}\right)=\mathrm{m}_{2} \mathrm{C}_{2}\left(\mathrm{~T}_{\mathrm{m}}-\mathrm{T}_{2}\right)$ $\mathrm{m} \times \mathrm{s} \times\left(\mathrm{T}-\mathrm{T}_{\mathrm{m}}\right)=\frac{\mathrm{m}}{2} \times 2 \mathrm{~s} \times\left(\mathrm{T}_{\mathrm{m}}-2 \mathrm{~T}\right)$ $\mathrm{T}-\mathrm{T}_{\mathrm{m}}=\mathrm{T}_{\mathrm{m}}-2 \mathrm{~T}$ $\mathrm{T}+2 \mathrm{~T}=2 \mathrm{~T}_{\mathrm{m}}$ $3 \mathrm{~T}=2 \mathrm{~T}_{\mathrm{m}}$ $\mathrm{T}_{\mathrm{m}}=\frac{3}{2} \mathrm{~T}$
148692
The change in the entropy of a 1 mole of an ideal gas which went through an isothermal process from an initial state $\left(\mathrm{P}_{1}, \mathrm{~V}_{1}, \mathrm{~T}\right)$ to the final state $\left(P_{2}, V_{2}, T\right)$ is equal to :
D Work done by $\mathrm{n}$ moles of the gas volume change from $V_{1}$ to $V_{2}$, $\mathrm{W}=\mathrm{nRT} \ln \frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}} \quad(\because \mathrm{n}=1)$ $\mathrm{W}=\mathrm{RT} \ln \frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}$ First law of thermodynamics, $\Delta \mathrm{Q}=\Delta \mathrm{U}+\mathrm{W} \quad \text { (for isothermal process, } \Delta \mathrm{U}=0)$ $\Delta \mathrm{Q}=\mathrm{W}$ From second law of thermodynamics, $\Delta S=\frac{\Delta Q}{T}=\frac{R T \ln \frac{V_{2}}{V_{1}}}{T}=R \ln \frac{V_{2}}{V_{1}}$
BCECE-2006
Thermodynamics
148693
If ' $\Delta Q$ ' is the amount of heat supplied to ' $n$ ' moles of a diatomic gas at constant pressure, ' $\Delta \mathbf{U}$ ' is the change in internal energy and ' $\Delta \mathbf{W}$ ' is the work done, then $\Delta W: \Delta \mathrm{U}: \Delta \mathbf{Q}$ is
1 $1: 2: 3$
2 $2: 5: 7$
3 $2: 3: 4$
4 $5: 7: 9$
Explanation:
B We know, At constant pressure $\Delta \mathrm{Q}=\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{T}$ $\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}$ $\mathrm{W}=\mathrm{P} \Delta \mathrm{V}=\mathrm{nR} \Delta \mathrm{T}$ $\therefore \quad \Delta \mathrm{Q}: \Delta \mathrm{U}: \mathrm{W}=\mathrm{C}_{\mathrm{p}}: \mathrm{C}_{\mathrm{v}}: \mathrm{R}$ For diatomic gas, $\mathrm{f}=5$ Gas is ideal and diatomic in nature. $\therefore \quad \mathrm{C}_{\mathrm{v}}=\frac{\mathrm{fR}}{2}=\frac{5 \mathrm{R}}{2}$ $\because \quad \mathrm{C}_{\mathrm{p}}=\mathrm{C}_{\mathrm{v}}+\mathrm{R}=\frac{5 \mathrm{R}}{2}+\mathrm{R}$ $=\frac{7 \mathrm{R}}{2}$ $\therefore \quad \Delta \mathrm{Q}: \Delta \mathrm{U}: \mathrm{W}=7: 5: 2$ Hence, $\mathrm{W}: \Delta \mathrm{U}: \Delta \mathrm{Q}=2: 5: 7$
MHT-CET 2020
Thermodynamics
148696
Find the ratio of $\frac{\Delta Q}{\Delta W}$ in an isobaric process, the ratio of molar specific heat capacities of the gas used is $\frac{C_{p}}{C_{v}}=\gamma$
1 $\frac{\gamma}{\gamma-1}$
2 $\frac{\gamma-1}{\gamma}$
3 $\frac{\gamma+1}{\gamma}$
4 $\frac{\gamma}{\gamma+1}$
Explanation:
A We know that, $\Delta \mathrm{Q}=\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{T}$ and $\quad \Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}$ Taking ratio of equation (i) and (ii), we get- $\frac{\Delta \mathrm{Q}}{\Delta \mathrm{U}}=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}$ $\because \quad \frac{\Delta \mathrm{Q}}{\Delta \mathrm{U}}=\gamma$ $\frac{\Delta \mathrm{Q}}{\Delta \mathrm{W}}=\frac{\Delta \mathrm{Q}}{\Delta \mathrm{Q}-\Delta \mathrm{U}}=\frac{1}{1-\frac{\Delta \mathrm{U}}{\Delta \mathrm{Q}}}$ $=\frac{1}{1-\frac{1}{\gamma}}$ $\frac{\Delta \mathrm{Q}}{\Delta \mathrm{W}}=\frac{\gamma}{\gamma-1}$
AP EAMCET-05.10.2021
Thermodynamics
148697
No process is possible whose sole result is the transfer of heat from a colder object to a hotter object. This is Clausius statement for
1 Zeroth law of thermodynamics
2 First law of thermodynamics
3 Second law of thermodynamics
4 Carnot's theorem
5 Principle of refrigeration
Explanation:
C There are two statements of second law of thermodynamics which are (1) Kelvin-Plank statement (2) Clausius statement Kelvin-Plank statement: It is impossible for a heat engine to produce a network in a complete cycle if it exchange heat only with a single reservoir. Clausius's statement: It is impossible to design a device, which works on a cycle and produces no effect other than heat transfer from a cold body to hot body.
Kerala CEE 04.07.2022
Thermodynamics
148700
A liquid of mass $m$ and specific heat $s$ is heated to a temperature $T$. Another liquid of mass $\mathrm{m} / \mathbf{2}$ and specific heat $2 \mathrm{~s}$ is heated to temperature 2T. If these two liquids are mixed, the resultant temperature of the mixture will be
1 $\frac{2}{3} \mathrm{~T}$
2 $\frac{8}{5} \mathrm{~T}$
3 $\frac{3}{4} \mathrm{~T}$
4 $\frac{3}{2} \mathrm{~T}$
Explanation:
D Given, Mass of liquid $\left(\mathrm{m}_{1}\right)=\mathrm{m}$ Specific heat of liquid $\left(\mathrm{C}_{1}\right)=\mathrm{s}$ Initial temperature of liquid $\left(\mathrm{T}_{1}\right)=\mathrm{T}$ Mass of liquid $\left(\mathrm{m}_{2}\right)=\mathrm{m} / 2$ Specific heat of liquid $\left(\mathrm{C}_{2}\right)=2 \mathrm{~s}$ Initial temperature of liquid $\left(\mathrm{T}_{2}\right)=2 \mathrm{~T}$ Let the resultant temperature of the mixture $=T_{m}$ Heat lost by liquid $1=$ Heat gained by liquid 2 $\mathrm{m}_{1} \mathrm{C}_{1}\left(\mathrm{~T}_{1}-\mathrm{T}_{\mathrm{m}}\right)=\mathrm{m}_{2} \mathrm{C}_{2}\left(\mathrm{~T}_{\mathrm{m}}-\mathrm{T}_{2}\right)$ $\mathrm{m} \times \mathrm{s} \times\left(\mathrm{T}-\mathrm{T}_{\mathrm{m}}\right)=\frac{\mathrm{m}}{2} \times 2 \mathrm{~s} \times\left(\mathrm{T}_{\mathrm{m}}-2 \mathrm{~T}\right)$ $\mathrm{T}-\mathrm{T}_{\mathrm{m}}=\mathrm{T}_{\mathrm{m}}-2 \mathrm{~T}$ $\mathrm{T}+2 \mathrm{~T}=2 \mathrm{~T}_{\mathrm{m}}$ $3 \mathrm{~T}=2 \mathrm{~T}_{\mathrm{m}}$ $\mathrm{T}_{\mathrm{m}}=\frac{3}{2} \mathrm{~T}$
148692
The change in the entropy of a 1 mole of an ideal gas which went through an isothermal process from an initial state $\left(\mathrm{P}_{1}, \mathrm{~V}_{1}, \mathrm{~T}\right)$ to the final state $\left(P_{2}, V_{2}, T\right)$ is equal to :
D Work done by $\mathrm{n}$ moles of the gas volume change from $V_{1}$ to $V_{2}$, $\mathrm{W}=\mathrm{nRT} \ln \frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}} \quad(\because \mathrm{n}=1)$ $\mathrm{W}=\mathrm{RT} \ln \frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}$ First law of thermodynamics, $\Delta \mathrm{Q}=\Delta \mathrm{U}+\mathrm{W} \quad \text { (for isothermal process, } \Delta \mathrm{U}=0)$ $\Delta \mathrm{Q}=\mathrm{W}$ From second law of thermodynamics, $\Delta S=\frac{\Delta Q}{T}=\frac{R T \ln \frac{V_{2}}{V_{1}}}{T}=R \ln \frac{V_{2}}{V_{1}}$
BCECE-2006
Thermodynamics
148693
If ' $\Delta Q$ ' is the amount of heat supplied to ' $n$ ' moles of a diatomic gas at constant pressure, ' $\Delta \mathbf{U}$ ' is the change in internal energy and ' $\Delta \mathbf{W}$ ' is the work done, then $\Delta W: \Delta \mathrm{U}: \Delta \mathbf{Q}$ is
1 $1: 2: 3$
2 $2: 5: 7$
3 $2: 3: 4$
4 $5: 7: 9$
Explanation:
B We know, At constant pressure $\Delta \mathrm{Q}=\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{T}$ $\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}$ $\mathrm{W}=\mathrm{P} \Delta \mathrm{V}=\mathrm{nR} \Delta \mathrm{T}$ $\therefore \quad \Delta \mathrm{Q}: \Delta \mathrm{U}: \mathrm{W}=\mathrm{C}_{\mathrm{p}}: \mathrm{C}_{\mathrm{v}}: \mathrm{R}$ For diatomic gas, $\mathrm{f}=5$ Gas is ideal and diatomic in nature. $\therefore \quad \mathrm{C}_{\mathrm{v}}=\frac{\mathrm{fR}}{2}=\frac{5 \mathrm{R}}{2}$ $\because \quad \mathrm{C}_{\mathrm{p}}=\mathrm{C}_{\mathrm{v}}+\mathrm{R}=\frac{5 \mathrm{R}}{2}+\mathrm{R}$ $=\frac{7 \mathrm{R}}{2}$ $\therefore \quad \Delta \mathrm{Q}: \Delta \mathrm{U}: \mathrm{W}=7: 5: 2$ Hence, $\mathrm{W}: \Delta \mathrm{U}: \Delta \mathrm{Q}=2: 5: 7$
MHT-CET 2020
Thermodynamics
148696
Find the ratio of $\frac{\Delta Q}{\Delta W}$ in an isobaric process, the ratio of molar specific heat capacities of the gas used is $\frac{C_{p}}{C_{v}}=\gamma$
1 $\frac{\gamma}{\gamma-1}$
2 $\frac{\gamma-1}{\gamma}$
3 $\frac{\gamma+1}{\gamma}$
4 $\frac{\gamma}{\gamma+1}$
Explanation:
A We know that, $\Delta \mathrm{Q}=\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{T}$ and $\quad \Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}$ Taking ratio of equation (i) and (ii), we get- $\frac{\Delta \mathrm{Q}}{\Delta \mathrm{U}}=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}$ $\because \quad \frac{\Delta \mathrm{Q}}{\Delta \mathrm{U}}=\gamma$ $\frac{\Delta \mathrm{Q}}{\Delta \mathrm{W}}=\frac{\Delta \mathrm{Q}}{\Delta \mathrm{Q}-\Delta \mathrm{U}}=\frac{1}{1-\frac{\Delta \mathrm{U}}{\Delta \mathrm{Q}}}$ $=\frac{1}{1-\frac{1}{\gamma}}$ $\frac{\Delta \mathrm{Q}}{\Delta \mathrm{W}}=\frac{\gamma}{\gamma-1}$
AP EAMCET-05.10.2021
Thermodynamics
148697
No process is possible whose sole result is the transfer of heat from a colder object to a hotter object. This is Clausius statement for
1 Zeroth law of thermodynamics
2 First law of thermodynamics
3 Second law of thermodynamics
4 Carnot's theorem
5 Principle of refrigeration
Explanation:
C There are two statements of second law of thermodynamics which are (1) Kelvin-Plank statement (2) Clausius statement Kelvin-Plank statement: It is impossible for a heat engine to produce a network in a complete cycle if it exchange heat only with a single reservoir. Clausius's statement: It is impossible to design a device, which works on a cycle and produces no effect other than heat transfer from a cold body to hot body.
Kerala CEE 04.07.2022
Thermodynamics
148700
A liquid of mass $m$ and specific heat $s$ is heated to a temperature $T$. Another liquid of mass $\mathrm{m} / \mathbf{2}$ and specific heat $2 \mathrm{~s}$ is heated to temperature 2T. If these two liquids are mixed, the resultant temperature of the mixture will be
1 $\frac{2}{3} \mathrm{~T}$
2 $\frac{8}{5} \mathrm{~T}$
3 $\frac{3}{4} \mathrm{~T}$
4 $\frac{3}{2} \mathrm{~T}$
Explanation:
D Given, Mass of liquid $\left(\mathrm{m}_{1}\right)=\mathrm{m}$ Specific heat of liquid $\left(\mathrm{C}_{1}\right)=\mathrm{s}$ Initial temperature of liquid $\left(\mathrm{T}_{1}\right)=\mathrm{T}$ Mass of liquid $\left(\mathrm{m}_{2}\right)=\mathrm{m} / 2$ Specific heat of liquid $\left(\mathrm{C}_{2}\right)=2 \mathrm{~s}$ Initial temperature of liquid $\left(\mathrm{T}_{2}\right)=2 \mathrm{~T}$ Let the resultant temperature of the mixture $=T_{m}$ Heat lost by liquid $1=$ Heat gained by liquid 2 $\mathrm{m}_{1} \mathrm{C}_{1}\left(\mathrm{~T}_{1}-\mathrm{T}_{\mathrm{m}}\right)=\mathrm{m}_{2} \mathrm{C}_{2}\left(\mathrm{~T}_{\mathrm{m}}-\mathrm{T}_{2}\right)$ $\mathrm{m} \times \mathrm{s} \times\left(\mathrm{T}-\mathrm{T}_{\mathrm{m}}\right)=\frac{\mathrm{m}}{2} \times 2 \mathrm{~s} \times\left(\mathrm{T}_{\mathrm{m}}-2 \mathrm{~T}\right)$ $\mathrm{T}-\mathrm{T}_{\mathrm{m}}=\mathrm{T}_{\mathrm{m}}-2 \mathrm{~T}$ $\mathrm{T}+2 \mathrm{~T}=2 \mathrm{~T}_{\mathrm{m}}$ $3 \mathrm{~T}=2 \mathrm{~T}_{\mathrm{m}}$ $\mathrm{T}_{\mathrm{m}}=\frac{3}{2} \mathrm{~T}$
148692
The change in the entropy of a 1 mole of an ideal gas which went through an isothermal process from an initial state $\left(\mathrm{P}_{1}, \mathrm{~V}_{1}, \mathrm{~T}\right)$ to the final state $\left(P_{2}, V_{2}, T\right)$ is equal to :
D Work done by $\mathrm{n}$ moles of the gas volume change from $V_{1}$ to $V_{2}$, $\mathrm{W}=\mathrm{nRT} \ln \frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}} \quad(\because \mathrm{n}=1)$ $\mathrm{W}=\mathrm{RT} \ln \frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}$ First law of thermodynamics, $\Delta \mathrm{Q}=\Delta \mathrm{U}+\mathrm{W} \quad \text { (for isothermal process, } \Delta \mathrm{U}=0)$ $\Delta \mathrm{Q}=\mathrm{W}$ From second law of thermodynamics, $\Delta S=\frac{\Delta Q}{T}=\frac{R T \ln \frac{V_{2}}{V_{1}}}{T}=R \ln \frac{V_{2}}{V_{1}}$
BCECE-2006
Thermodynamics
148693
If ' $\Delta Q$ ' is the amount of heat supplied to ' $n$ ' moles of a diatomic gas at constant pressure, ' $\Delta \mathbf{U}$ ' is the change in internal energy and ' $\Delta \mathbf{W}$ ' is the work done, then $\Delta W: \Delta \mathrm{U}: \Delta \mathbf{Q}$ is
1 $1: 2: 3$
2 $2: 5: 7$
3 $2: 3: 4$
4 $5: 7: 9$
Explanation:
B We know, At constant pressure $\Delta \mathrm{Q}=\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{T}$ $\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}$ $\mathrm{W}=\mathrm{P} \Delta \mathrm{V}=\mathrm{nR} \Delta \mathrm{T}$ $\therefore \quad \Delta \mathrm{Q}: \Delta \mathrm{U}: \mathrm{W}=\mathrm{C}_{\mathrm{p}}: \mathrm{C}_{\mathrm{v}}: \mathrm{R}$ For diatomic gas, $\mathrm{f}=5$ Gas is ideal and diatomic in nature. $\therefore \quad \mathrm{C}_{\mathrm{v}}=\frac{\mathrm{fR}}{2}=\frac{5 \mathrm{R}}{2}$ $\because \quad \mathrm{C}_{\mathrm{p}}=\mathrm{C}_{\mathrm{v}}+\mathrm{R}=\frac{5 \mathrm{R}}{2}+\mathrm{R}$ $=\frac{7 \mathrm{R}}{2}$ $\therefore \quad \Delta \mathrm{Q}: \Delta \mathrm{U}: \mathrm{W}=7: 5: 2$ Hence, $\mathrm{W}: \Delta \mathrm{U}: \Delta \mathrm{Q}=2: 5: 7$
MHT-CET 2020
Thermodynamics
148696
Find the ratio of $\frac{\Delta Q}{\Delta W}$ in an isobaric process, the ratio of molar specific heat capacities of the gas used is $\frac{C_{p}}{C_{v}}=\gamma$
1 $\frac{\gamma}{\gamma-1}$
2 $\frac{\gamma-1}{\gamma}$
3 $\frac{\gamma+1}{\gamma}$
4 $\frac{\gamma}{\gamma+1}$
Explanation:
A We know that, $\Delta \mathrm{Q}=\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{T}$ and $\quad \Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}$ Taking ratio of equation (i) and (ii), we get- $\frac{\Delta \mathrm{Q}}{\Delta \mathrm{U}}=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}$ $\because \quad \frac{\Delta \mathrm{Q}}{\Delta \mathrm{U}}=\gamma$ $\frac{\Delta \mathrm{Q}}{\Delta \mathrm{W}}=\frac{\Delta \mathrm{Q}}{\Delta \mathrm{Q}-\Delta \mathrm{U}}=\frac{1}{1-\frac{\Delta \mathrm{U}}{\Delta \mathrm{Q}}}$ $=\frac{1}{1-\frac{1}{\gamma}}$ $\frac{\Delta \mathrm{Q}}{\Delta \mathrm{W}}=\frac{\gamma}{\gamma-1}$
AP EAMCET-05.10.2021
Thermodynamics
148697
No process is possible whose sole result is the transfer of heat from a colder object to a hotter object. This is Clausius statement for
1 Zeroth law of thermodynamics
2 First law of thermodynamics
3 Second law of thermodynamics
4 Carnot's theorem
5 Principle of refrigeration
Explanation:
C There are two statements of second law of thermodynamics which are (1) Kelvin-Plank statement (2) Clausius statement Kelvin-Plank statement: It is impossible for a heat engine to produce a network in a complete cycle if it exchange heat only with a single reservoir. Clausius's statement: It is impossible to design a device, which works on a cycle and produces no effect other than heat transfer from a cold body to hot body.
Kerala CEE 04.07.2022
Thermodynamics
148700
A liquid of mass $m$ and specific heat $s$ is heated to a temperature $T$. Another liquid of mass $\mathrm{m} / \mathbf{2}$ and specific heat $2 \mathrm{~s}$ is heated to temperature 2T. If these two liquids are mixed, the resultant temperature of the mixture will be
1 $\frac{2}{3} \mathrm{~T}$
2 $\frac{8}{5} \mathrm{~T}$
3 $\frac{3}{4} \mathrm{~T}$
4 $\frac{3}{2} \mathrm{~T}$
Explanation:
D Given, Mass of liquid $\left(\mathrm{m}_{1}\right)=\mathrm{m}$ Specific heat of liquid $\left(\mathrm{C}_{1}\right)=\mathrm{s}$ Initial temperature of liquid $\left(\mathrm{T}_{1}\right)=\mathrm{T}$ Mass of liquid $\left(\mathrm{m}_{2}\right)=\mathrm{m} / 2$ Specific heat of liquid $\left(\mathrm{C}_{2}\right)=2 \mathrm{~s}$ Initial temperature of liquid $\left(\mathrm{T}_{2}\right)=2 \mathrm{~T}$ Let the resultant temperature of the mixture $=T_{m}$ Heat lost by liquid $1=$ Heat gained by liquid 2 $\mathrm{m}_{1} \mathrm{C}_{1}\left(\mathrm{~T}_{1}-\mathrm{T}_{\mathrm{m}}\right)=\mathrm{m}_{2} \mathrm{C}_{2}\left(\mathrm{~T}_{\mathrm{m}}-\mathrm{T}_{2}\right)$ $\mathrm{m} \times \mathrm{s} \times\left(\mathrm{T}-\mathrm{T}_{\mathrm{m}}\right)=\frac{\mathrm{m}}{2} \times 2 \mathrm{~s} \times\left(\mathrm{T}_{\mathrm{m}}-2 \mathrm{~T}\right)$ $\mathrm{T}-\mathrm{T}_{\mathrm{m}}=\mathrm{T}_{\mathrm{m}}-2 \mathrm{~T}$ $\mathrm{T}+2 \mathrm{~T}=2 \mathrm{~T}_{\mathrm{m}}$ $3 \mathrm{~T}=2 \mathrm{~T}_{\mathrm{m}}$ $\mathrm{T}_{\mathrm{m}}=\frac{3}{2} \mathrm{~T}$
