148445
Assertion: Adiabatic expansion is always accompanied by fall in temperature. Reason: In adiabatic process, volume is inversely proportional to temperature.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
C According to option the equation of state for an adiabatic process is- $\mathrm{TV}^{\gamma-1}=\text { constant }$ $\mathrm{T}=\frac{\text { constant }}{\mathrm{V}^{\gamma-1}}$ So, if the gas is expanded then temperature falls. But the temperature for the process is inversely proportional to $\mathrm{V}^{\gamma-1}$.
AIIMS-2011
Thermodynamics
148448
An ideal gas at $27^{\circ} \mathrm{C}$ is compressed adiabatically to $(8 / 27)$ of its original volume. If ratio of specific heats, $\gamma=5 / 3$ then the rise in temperature of the gas is $375 \mathrm{~K}$
1 $500 \mathrm{~K}$
2 $125 \mathrm{~K}$
3 $250 \mathrm{~K}$
4 $375 \mathrm{~K}$
Explanation:
D Given, $\mathrm{T}_{1}=27^{\circ} \mathrm{C}=300 \mathrm{~K}$ $\mathrm{~V}_{1}=\mathrm{V}$ $\mathrm{V}_{2}=\frac{8}{27} \mathrm{~V}$ $\gamma=\frac{5}{3}$ For adiabatic process, $\mathrm{TV}^{\gamma-1}=\text { constant }$ So, $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)^{\gamma-1}$ Put the given value in equation (i) $\frac{300}{\mathrm{~T}_{2}}=\left(\frac{\left(\frac{8}{27} \mathrm{~V}\right)}{\mathrm{V}}\right)^{\frac{5}{3}-1}$ $\mathrm{~T}_{2}=300 \times \frac{9}{4}$ $\mathrm{~T}_{2}=675 \mathrm{~K}$ Rise in temperature $(\Delta T)=675-300$ Rise in temperature $(\Delta \mathrm{T})=375 \mathrm{~K}$
UPCPMT 1984
Thermodynamics
148450
A gas expands adiabatically at constant pressure, such that its temperature $T \propto \frac{1}{\sqrt{V}}$. The value of $\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{v}}$ of the gas is
1 1.30
2 1.50
3 1.67
4 2.00
Explanation:
B $\mathrm{TV}^{\gamma-1}=$ Constant $\frac{\mathrm{K}}{\mathrm{R}}=$ Constant $\mathrm{T} \propto \frac{1}{\sqrt{\mathrm{V}}}$ $\mathrm{T}=\mathrm{K} \frac{1}{\sqrt{\mathrm{V}}} \quad(\because \mathrm{K}=$ proportionality constant $)$ $\mathrm{TV}^{\frac{1}{2}}=\mathrm{K}$ $\mathrm{TV}^{\frac{1}{2}}=$ constant By comparing equation (i) and (ii) we get $\gamma-1=\frac{1}{2}$ $\gamma=\frac{1}{2}+1$ $\gamma=\frac{3}{2}$ $=1.5$
MHT-CET 2004
Thermodynamics
148451
A sample of gas with $\gamma=1.5$ is taken through an adiabatic process in which the volume is compressed from $1600 \mathrm{~cm}^{3}$ to $400 \mathrm{~cm}^{3}$ If the initial pressure is $150 \mathrm{kPa}$, then work done in the process is
1 $-120 \mathrm{~J}$
2 $+120 \mathrm{~J}$
3 $-480 \mathrm{~J}$
4 $+480 \mathrm{~J}$
Explanation:
C Given that, $\mathrm{P}_{1}=150 \mathrm{kPa}, \mathrm{V}_{1}=1600 \mathrm{~cm}^{3}, \mathrm{~V}_{2}=400 \mathrm{~cm}^{3}$ and $\gamma=1.5$ As the process is adiabatic, So, by using adiabatic equation, i.e. $P_{1} V_{1}^{\gamma}=P_{2} V_{2}^{\gamma}$ $P_{2}=P_{1}\left(\frac{V_{1}}{V_{2}}\right)^{\gamma}$ $=150\left(\frac{600}{400}\right)^{1.5}$ $=150 \times 4^{1.5}$ $=1200 \mathrm{kPa}$ Work done in adiabatic process $\mathrm{W}=\frac{\mathrm{P}_{1} \mathrm{~V}_{1}-\mathrm{P}_{2} \mathrm{~V}_{2}}{\gamma-1}$ $\mathrm{~W}=\frac{150 \times 1600-1200 \times 400}{1.5-1}$ $\mathrm{~W}=\frac{240-480}{0.5}$ $\mathrm{~W}=-480 \mathrm{~J}$
148445
Assertion: Adiabatic expansion is always accompanied by fall in temperature. Reason: In adiabatic process, volume is inversely proportional to temperature.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
C According to option the equation of state for an adiabatic process is- $\mathrm{TV}^{\gamma-1}=\text { constant }$ $\mathrm{T}=\frac{\text { constant }}{\mathrm{V}^{\gamma-1}}$ So, if the gas is expanded then temperature falls. But the temperature for the process is inversely proportional to $\mathrm{V}^{\gamma-1}$.
AIIMS-2011
Thermodynamics
148448
An ideal gas at $27^{\circ} \mathrm{C}$ is compressed adiabatically to $(8 / 27)$ of its original volume. If ratio of specific heats, $\gamma=5 / 3$ then the rise in temperature of the gas is $375 \mathrm{~K}$
1 $500 \mathrm{~K}$
2 $125 \mathrm{~K}$
3 $250 \mathrm{~K}$
4 $375 \mathrm{~K}$
Explanation:
D Given, $\mathrm{T}_{1}=27^{\circ} \mathrm{C}=300 \mathrm{~K}$ $\mathrm{~V}_{1}=\mathrm{V}$ $\mathrm{V}_{2}=\frac{8}{27} \mathrm{~V}$ $\gamma=\frac{5}{3}$ For adiabatic process, $\mathrm{TV}^{\gamma-1}=\text { constant }$ So, $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)^{\gamma-1}$ Put the given value in equation (i) $\frac{300}{\mathrm{~T}_{2}}=\left(\frac{\left(\frac{8}{27} \mathrm{~V}\right)}{\mathrm{V}}\right)^{\frac{5}{3}-1}$ $\mathrm{~T}_{2}=300 \times \frac{9}{4}$ $\mathrm{~T}_{2}=675 \mathrm{~K}$ Rise in temperature $(\Delta T)=675-300$ Rise in temperature $(\Delta \mathrm{T})=375 \mathrm{~K}$
UPCPMT 1984
Thermodynamics
148450
A gas expands adiabatically at constant pressure, such that its temperature $T \propto \frac{1}{\sqrt{V}}$. The value of $\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{v}}$ of the gas is
1 1.30
2 1.50
3 1.67
4 2.00
Explanation:
B $\mathrm{TV}^{\gamma-1}=$ Constant $\frac{\mathrm{K}}{\mathrm{R}}=$ Constant $\mathrm{T} \propto \frac{1}{\sqrt{\mathrm{V}}}$ $\mathrm{T}=\mathrm{K} \frac{1}{\sqrt{\mathrm{V}}} \quad(\because \mathrm{K}=$ proportionality constant $)$ $\mathrm{TV}^{\frac{1}{2}}=\mathrm{K}$ $\mathrm{TV}^{\frac{1}{2}}=$ constant By comparing equation (i) and (ii) we get $\gamma-1=\frac{1}{2}$ $\gamma=\frac{1}{2}+1$ $\gamma=\frac{3}{2}$ $=1.5$
MHT-CET 2004
Thermodynamics
148451
A sample of gas with $\gamma=1.5$ is taken through an adiabatic process in which the volume is compressed from $1600 \mathrm{~cm}^{3}$ to $400 \mathrm{~cm}^{3}$ If the initial pressure is $150 \mathrm{kPa}$, then work done in the process is
1 $-120 \mathrm{~J}$
2 $+120 \mathrm{~J}$
3 $-480 \mathrm{~J}$
4 $+480 \mathrm{~J}$
Explanation:
C Given that, $\mathrm{P}_{1}=150 \mathrm{kPa}, \mathrm{V}_{1}=1600 \mathrm{~cm}^{3}, \mathrm{~V}_{2}=400 \mathrm{~cm}^{3}$ and $\gamma=1.5$ As the process is adiabatic, So, by using adiabatic equation, i.e. $P_{1} V_{1}^{\gamma}=P_{2} V_{2}^{\gamma}$ $P_{2}=P_{1}\left(\frac{V_{1}}{V_{2}}\right)^{\gamma}$ $=150\left(\frac{600}{400}\right)^{1.5}$ $=150 \times 4^{1.5}$ $=1200 \mathrm{kPa}$ Work done in adiabatic process $\mathrm{W}=\frac{\mathrm{P}_{1} \mathrm{~V}_{1}-\mathrm{P}_{2} \mathrm{~V}_{2}}{\gamma-1}$ $\mathrm{~W}=\frac{150 \times 1600-1200 \times 400}{1.5-1}$ $\mathrm{~W}=\frac{240-480}{0.5}$ $\mathrm{~W}=-480 \mathrm{~J}$
148445
Assertion: Adiabatic expansion is always accompanied by fall in temperature. Reason: In adiabatic process, volume is inversely proportional to temperature.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
C According to option the equation of state for an adiabatic process is- $\mathrm{TV}^{\gamma-1}=\text { constant }$ $\mathrm{T}=\frac{\text { constant }}{\mathrm{V}^{\gamma-1}}$ So, if the gas is expanded then temperature falls. But the temperature for the process is inversely proportional to $\mathrm{V}^{\gamma-1}$.
AIIMS-2011
Thermodynamics
148448
An ideal gas at $27^{\circ} \mathrm{C}$ is compressed adiabatically to $(8 / 27)$ of its original volume. If ratio of specific heats, $\gamma=5 / 3$ then the rise in temperature of the gas is $375 \mathrm{~K}$
1 $500 \mathrm{~K}$
2 $125 \mathrm{~K}$
3 $250 \mathrm{~K}$
4 $375 \mathrm{~K}$
Explanation:
D Given, $\mathrm{T}_{1}=27^{\circ} \mathrm{C}=300 \mathrm{~K}$ $\mathrm{~V}_{1}=\mathrm{V}$ $\mathrm{V}_{2}=\frac{8}{27} \mathrm{~V}$ $\gamma=\frac{5}{3}$ For adiabatic process, $\mathrm{TV}^{\gamma-1}=\text { constant }$ So, $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)^{\gamma-1}$ Put the given value in equation (i) $\frac{300}{\mathrm{~T}_{2}}=\left(\frac{\left(\frac{8}{27} \mathrm{~V}\right)}{\mathrm{V}}\right)^{\frac{5}{3}-1}$ $\mathrm{~T}_{2}=300 \times \frac{9}{4}$ $\mathrm{~T}_{2}=675 \mathrm{~K}$ Rise in temperature $(\Delta T)=675-300$ Rise in temperature $(\Delta \mathrm{T})=375 \mathrm{~K}$
UPCPMT 1984
Thermodynamics
148450
A gas expands adiabatically at constant pressure, such that its temperature $T \propto \frac{1}{\sqrt{V}}$. The value of $\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{v}}$ of the gas is
1 1.30
2 1.50
3 1.67
4 2.00
Explanation:
B $\mathrm{TV}^{\gamma-1}=$ Constant $\frac{\mathrm{K}}{\mathrm{R}}=$ Constant $\mathrm{T} \propto \frac{1}{\sqrt{\mathrm{V}}}$ $\mathrm{T}=\mathrm{K} \frac{1}{\sqrt{\mathrm{V}}} \quad(\because \mathrm{K}=$ proportionality constant $)$ $\mathrm{TV}^{\frac{1}{2}}=\mathrm{K}$ $\mathrm{TV}^{\frac{1}{2}}=$ constant By comparing equation (i) and (ii) we get $\gamma-1=\frac{1}{2}$ $\gamma=\frac{1}{2}+1$ $\gamma=\frac{3}{2}$ $=1.5$
MHT-CET 2004
Thermodynamics
148451
A sample of gas with $\gamma=1.5$ is taken through an adiabatic process in which the volume is compressed from $1600 \mathrm{~cm}^{3}$ to $400 \mathrm{~cm}^{3}$ If the initial pressure is $150 \mathrm{kPa}$, then work done in the process is
1 $-120 \mathrm{~J}$
2 $+120 \mathrm{~J}$
3 $-480 \mathrm{~J}$
4 $+480 \mathrm{~J}$
Explanation:
C Given that, $\mathrm{P}_{1}=150 \mathrm{kPa}, \mathrm{V}_{1}=1600 \mathrm{~cm}^{3}, \mathrm{~V}_{2}=400 \mathrm{~cm}^{3}$ and $\gamma=1.5$ As the process is adiabatic, So, by using adiabatic equation, i.e. $P_{1} V_{1}^{\gamma}=P_{2} V_{2}^{\gamma}$ $P_{2}=P_{1}\left(\frac{V_{1}}{V_{2}}\right)^{\gamma}$ $=150\left(\frac{600}{400}\right)^{1.5}$ $=150 \times 4^{1.5}$ $=1200 \mathrm{kPa}$ Work done in adiabatic process $\mathrm{W}=\frac{\mathrm{P}_{1} \mathrm{~V}_{1}-\mathrm{P}_{2} \mathrm{~V}_{2}}{\gamma-1}$ $\mathrm{~W}=\frac{150 \times 1600-1200 \times 400}{1.5-1}$ $\mathrm{~W}=\frac{240-480}{0.5}$ $\mathrm{~W}=-480 \mathrm{~J}$
148445
Assertion: Adiabatic expansion is always accompanied by fall in temperature. Reason: In adiabatic process, volume is inversely proportional to temperature.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
C According to option the equation of state for an adiabatic process is- $\mathrm{TV}^{\gamma-1}=\text { constant }$ $\mathrm{T}=\frac{\text { constant }}{\mathrm{V}^{\gamma-1}}$ So, if the gas is expanded then temperature falls. But the temperature for the process is inversely proportional to $\mathrm{V}^{\gamma-1}$.
AIIMS-2011
Thermodynamics
148448
An ideal gas at $27^{\circ} \mathrm{C}$ is compressed adiabatically to $(8 / 27)$ of its original volume. If ratio of specific heats, $\gamma=5 / 3$ then the rise in temperature of the gas is $375 \mathrm{~K}$
1 $500 \mathrm{~K}$
2 $125 \mathrm{~K}$
3 $250 \mathrm{~K}$
4 $375 \mathrm{~K}$
Explanation:
D Given, $\mathrm{T}_{1}=27^{\circ} \mathrm{C}=300 \mathrm{~K}$ $\mathrm{~V}_{1}=\mathrm{V}$ $\mathrm{V}_{2}=\frac{8}{27} \mathrm{~V}$ $\gamma=\frac{5}{3}$ For adiabatic process, $\mathrm{TV}^{\gamma-1}=\text { constant }$ So, $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)^{\gamma-1}$ Put the given value in equation (i) $\frac{300}{\mathrm{~T}_{2}}=\left(\frac{\left(\frac{8}{27} \mathrm{~V}\right)}{\mathrm{V}}\right)^{\frac{5}{3}-1}$ $\mathrm{~T}_{2}=300 \times \frac{9}{4}$ $\mathrm{~T}_{2}=675 \mathrm{~K}$ Rise in temperature $(\Delta T)=675-300$ Rise in temperature $(\Delta \mathrm{T})=375 \mathrm{~K}$
UPCPMT 1984
Thermodynamics
148450
A gas expands adiabatically at constant pressure, such that its temperature $T \propto \frac{1}{\sqrt{V}}$. The value of $\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{v}}$ of the gas is
1 1.30
2 1.50
3 1.67
4 2.00
Explanation:
B $\mathrm{TV}^{\gamma-1}=$ Constant $\frac{\mathrm{K}}{\mathrm{R}}=$ Constant $\mathrm{T} \propto \frac{1}{\sqrt{\mathrm{V}}}$ $\mathrm{T}=\mathrm{K} \frac{1}{\sqrt{\mathrm{V}}} \quad(\because \mathrm{K}=$ proportionality constant $)$ $\mathrm{TV}^{\frac{1}{2}}=\mathrm{K}$ $\mathrm{TV}^{\frac{1}{2}}=$ constant By comparing equation (i) and (ii) we get $\gamma-1=\frac{1}{2}$ $\gamma=\frac{1}{2}+1$ $\gamma=\frac{3}{2}$ $=1.5$
MHT-CET 2004
Thermodynamics
148451
A sample of gas with $\gamma=1.5$ is taken through an adiabatic process in which the volume is compressed from $1600 \mathrm{~cm}^{3}$ to $400 \mathrm{~cm}^{3}$ If the initial pressure is $150 \mathrm{kPa}$, then work done in the process is
1 $-120 \mathrm{~J}$
2 $+120 \mathrm{~J}$
3 $-480 \mathrm{~J}$
4 $+480 \mathrm{~J}$
Explanation:
C Given that, $\mathrm{P}_{1}=150 \mathrm{kPa}, \mathrm{V}_{1}=1600 \mathrm{~cm}^{3}, \mathrm{~V}_{2}=400 \mathrm{~cm}^{3}$ and $\gamma=1.5$ As the process is adiabatic, So, by using adiabatic equation, i.e. $P_{1} V_{1}^{\gamma}=P_{2} V_{2}^{\gamma}$ $P_{2}=P_{1}\left(\frac{V_{1}}{V_{2}}\right)^{\gamma}$ $=150\left(\frac{600}{400}\right)^{1.5}$ $=150 \times 4^{1.5}$ $=1200 \mathrm{kPa}$ Work done in adiabatic process $\mathrm{W}=\frac{\mathrm{P}_{1} \mathrm{~V}_{1}-\mathrm{P}_{2} \mathrm{~V}_{2}}{\gamma-1}$ $\mathrm{~W}=\frac{150 \times 1600-1200 \times 400}{1.5-1}$ $\mathrm{~W}=\frac{240-480}{0.5}$ $\mathrm{~W}=-480 \mathrm{~J}$
