138938
A monoatomic gas $\left(\gamma=\frac{5}{3}\right)$ is suddenly compressed to $1 / 8^{\text {th }}$ of the original volume adiabatically, then the pressure of the gas will become how many times larger than pressure originally?
1 32
2 $40 / 3$
3 8
4 same as before
Explanation:
A Given that, For monoatomic gas, $\gamma=\frac{5}{3}$ As we know that, for adiabatic process $\mathrm{PV}^{\gamma}=\text { constant }$ $\mathrm{P}_{1} \mathrm{~V}_{1}^{\gamma}=\mathrm{P}_{2} \mathrm{~V}_{2}^{\gamma}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\left(\frac{8 \mathrm{~V}_{1}}{\mathrm{~V}_{1}}\right)^{5 / 3}=\left(2^{3}\right)^{5 / 3}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=2^{5}=32$ $\mathrm{P}_{2}=32 \mathrm{P}_{1}$
J and K CET- 2000
Kinetic Theory of Gases
138940
An ideal gas undergoes a process that has the net effect of doubling its temperature and doubling its pressure. If $V$ was the initial volume of the gas, the final volume is
1 $\mathrm{V}$
2 $2 \mathrm{~V}$
3 $4 \mathrm{~V}$
4 $\frac{\mathrm{V}}{4}$
Explanation:
A For an ideal gas equation, $\quad \mathrm{PV}=\mathrm{nRT}$ $\therefore \quad \frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{P}_{2} \mathrm{~V}_{2}}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{2 \mathrm{P}_{1} \mathrm{~V}_{2}}=\frac{\mathrm{T}_{1}}{2 \mathrm{~T}_{1}} \quad\left[\because \mathrm{P}_{2}=2 \mathrm{P}_{1}, \mathrm{~T}_{2}=2 \mathrm{~T}_{1}\right]$ $\frac{\mathrm{V}_{1}}{2 \mathrm{~V}_{2}}=\frac{\mathrm{T}_{1}}{2 \mathrm{~T}_{1}}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{1}{1} \Rightarrow \mathrm{V}_{2}=\mathrm{V}_{1}=\mathrm{V} \quad\left(\because \mathrm{V}_{1}=\mathrm{V}\right)$
J and K CET- 1997
Kinetic Theory of Gases
138943
The volume of a gas at $20^{\circ} \mathrm{C}$ is $100 \mathrm{~cm}^{3}$ at 1 atmospheric pressure. If it is heated to $100^{\circ} \mathrm{C}$, its volume becomes $125 \mathrm{~cm}^{3}$ at the same pressure, then coefficient of gas at constant pressure is
1 $3.6 \times 10^{-3} /{ }^{\circ} \mathrm{C}$
2 $3.6 \times 10^{-4} /{ }^{\circ} \mathrm{C}$
3 $2.5 \times 10^{-3} /{ }^{\circ} \mathrm{C}$
4 $1.25 \times 10^{-3} /{ }^{\circ} \mathrm{C}$
Explanation:
A Given, Initial volume of a gas $\left(\mathrm{V}_{\mathrm{i}}\right)=100 \mathrm{~cm}^{3}$ Initial temperature of a gas $\left(\mathrm{T}_{\mathrm{i}}\right)=20^{\circ} \mathrm{C}$ Final volume of a gas $\left(\mathrm{V}_{\mathrm{f}}\right)=125 \mathrm{~cm}^{3}$ Final temperature of gas $\left(\mathrm{T}_{\mathrm{f}}\right)=100^{\circ} \mathrm{C}$ We know that, $\mathrm{V}_{\mathrm{f}}=\mathrm{V}_{\mathrm{i}}(1+\gamma \Delta \mathrm{T})$ $125=100[1+\gamma \times(100-20)]$ $1+\gamma(80)=\frac{125}{100}$ $1+80 \times \gamma=\frac{5}{4}$ $80 \times \gamma=\frac{5}{4}-1$ $80 \times \gamma=\frac{1}{4}$ $\gamma =\frac{1}{80 \times 4}$ $\gamma =3.125 \times 10^{-3} /{ }^{\circ} \mathrm{C}$ $\gamma \approx 3.6 \times 10^{-3} /{ }^{\circ} \mathrm{C}$
UP CPMT-2001
Kinetic Theory of Gases
138945
1 mole of an ideal gas at an initial temperature of $T K$ does $6 R$ joule of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is $5 / 3$, then final temperature of the gas will be
1 $(\mathrm{T}-4) \mathrm{K}$
2 $(\mathrm{T}+4) \mathrm{K}$
3 $(\mathrm{T}-2.4) \mathrm{K}$
4 $(\mathrm{T}+2.4) \mathrm{K}$
Explanation:
A Given, Initial temperature $=\mathrm{TK}$ Adiabatic work $=6 \mathrm{R}$ Joule $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma=5 / 3$ We know that, $\text { Work done }=\frac{\mathrm{P}_{1} \mathrm{~V}_{1}-\mathrm{P}_{2} \mathrm{~V}_{2}}{\gamma-1}$ $=\frac{\mathrm{nRT}_{1}-\mathrm{nRT}_{2}}{(\gamma-1)}$ $[\because \mathrm{PV}=\mathrm{nRT} \text{ Ideal gas equation}]$ $6 \mathrm{R}=\frac{\mathrm{nR}\left[\mathrm{T}_{1}-\mathrm{T}_{2}\right]}{\gamma-1}$ $6=\frac{\mathrm{T}-\mathrm{T}_{2}}{5 / 3-1}$ $6=\frac{3}{2}\left[\mathrm{~T}-\mathrm{T}_{2}\right]$ $4=\mathrm{T}-\mathrm{T}_{2}$ $\therefore \quad \mathrm{T}_{2}=(\mathrm{T}-4) \mathrm{K}$
138938
A monoatomic gas $\left(\gamma=\frac{5}{3}\right)$ is suddenly compressed to $1 / 8^{\text {th }}$ of the original volume adiabatically, then the pressure of the gas will become how many times larger than pressure originally?
1 32
2 $40 / 3$
3 8
4 same as before
Explanation:
A Given that, For monoatomic gas, $\gamma=\frac{5}{3}$ As we know that, for adiabatic process $\mathrm{PV}^{\gamma}=\text { constant }$ $\mathrm{P}_{1} \mathrm{~V}_{1}^{\gamma}=\mathrm{P}_{2} \mathrm{~V}_{2}^{\gamma}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\left(\frac{8 \mathrm{~V}_{1}}{\mathrm{~V}_{1}}\right)^{5 / 3}=\left(2^{3}\right)^{5 / 3}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=2^{5}=32$ $\mathrm{P}_{2}=32 \mathrm{P}_{1}$
J and K CET- 2000
Kinetic Theory of Gases
138940
An ideal gas undergoes a process that has the net effect of doubling its temperature and doubling its pressure. If $V$ was the initial volume of the gas, the final volume is
1 $\mathrm{V}$
2 $2 \mathrm{~V}$
3 $4 \mathrm{~V}$
4 $\frac{\mathrm{V}}{4}$
Explanation:
A For an ideal gas equation, $\quad \mathrm{PV}=\mathrm{nRT}$ $\therefore \quad \frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{P}_{2} \mathrm{~V}_{2}}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{2 \mathrm{P}_{1} \mathrm{~V}_{2}}=\frac{\mathrm{T}_{1}}{2 \mathrm{~T}_{1}} \quad\left[\because \mathrm{P}_{2}=2 \mathrm{P}_{1}, \mathrm{~T}_{2}=2 \mathrm{~T}_{1}\right]$ $\frac{\mathrm{V}_{1}}{2 \mathrm{~V}_{2}}=\frac{\mathrm{T}_{1}}{2 \mathrm{~T}_{1}}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{1}{1} \Rightarrow \mathrm{V}_{2}=\mathrm{V}_{1}=\mathrm{V} \quad\left(\because \mathrm{V}_{1}=\mathrm{V}\right)$
J and K CET- 1997
Kinetic Theory of Gases
138943
The volume of a gas at $20^{\circ} \mathrm{C}$ is $100 \mathrm{~cm}^{3}$ at 1 atmospheric pressure. If it is heated to $100^{\circ} \mathrm{C}$, its volume becomes $125 \mathrm{~cm}^{3}$ at the same pressure, then coefficient of gas at constant pressure is
1 $3.6 \times 10^{-3} /{ }^{\circ} \mathrm{C}$
2 $3.6 \times 10^{-4} /{ }^{\circ} \mathrm{C}$
3 $2.5 \times 10^{-3} /{ }^{\circ} \mathrm{C}$
4 $1.25 \times 10^{-3} /{ }^{\circ} \mathrm{C}$
Explanation:
A Given, Initial volume of a gas $\left(\mathrm{V}_{\mathrm{i}}\right)=100 \mathrm{~cm}^{3}$ Initial temperature of a gas $\left(\mathrm{T}_{\mathrm{i}}\right)=20^{\circ} \mathrm{C}$ Final volume of a gas $\left(\mathrm{V}_{\mathrm{f}}\right)=125 \mathrm{~cm}^{3}$ Final temperature of gas $\left(\mathrm{T}_{\mathrm{f}}\right)=100^{\circ} \mathrm{C}$ We know that, $\mathrm{V}_{\mathrm{f}}=\mathrm{V}_{\mathrm{i}}(1+\gamma \Delta \mathrm{T})$ $125=100[1+\gamma \times(100-20)]$ $1+\gamma(80)=\frac{125}{100}$ $1+80 \times \gamma=\frac{5}{4}$ $80 \times \gamma=\frac{5}{4}-1$ $80 \times \gamma=\frac{1}{4}$ $\gamma =\frac{1}{80 \times 4}$ $\gamma =3.125 \times 10^{-3} /{ }^{\circ} \mathrm{C}$ $\gamma \approx 3.6 \times 10^{-3} /{ }^{\circ} \mathrm{C}$
UP CPMT-2001
Kinetic Theory of Gases
138945
1 mole of an ideal gas at an initial temperature of $T K$ does $6 R$ joule of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is $5 / 3$, then final temperature of the gas will be
1 $(\mathrm{T}-4) \mathrm{K}$
2 $(\mathrm{T}+4) \mathrm{K}$
3 $(\mathrm{T}-2.4) \mathrm{K}$
4 $(\mathrm{T}+2.4) \mathrm{K}$
Explanation:
A Given, Initial temperature $=\mathrm{TK}$ Adiabatic work $=6 \mathrm{R}$ Joule $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma=5 / 3$ We know that, $\text { Work done }=\frac{\mathrm{P}_{1} \mathrm{~V}_{1}-\mathrm{P}_{2} \mathrm{~V}_{2}}{\gamma-1}$ $=\frac{\mathrm{nRT}_{1}-\mathrm{nRT}_{2}}{(\gamma-1)}$ $[\because \mathrm{PV}=\mathrm{nRT} \text{ Ideal gas equation}]$ $6 \mathrm{R}=\frac{\mathrm{nR}\left[\mathrm{T}_{1}-\mathrm{T}_{2}\right]}{\gamma-1}$ $6=\frac{\mathrm{T}-\mathrm{T}_{2}}{5 / 3-1}$ $6=\frac{3}{2}\left[\mathrm{~T}-\mathrm{T}_{2}\right]$ $4=\mathrm{T}-\mathrm{T}_{2}$ $\therefore \quad \mathrm{T}_{2}=(\mathrm{T}-4) \mathrm{K}$
138938
A monoatomic gas $\left(\gamma=\frac{5}{3}\right)$ is suddenly compressed to $1 / 8^{\text {th }}$ of the original volume adiabatically, then the pressure of the gas will become how many times larger than pressure originally?
1 32
2 $40 / 3$
3 8
4 same as before
Explanation:
A Given that, For monoatomic gas, $\gamma=\frac{5}{3}$ As we know that, for adiabatic process $\mathrm{PV}^{\gamma}=\text { constant }$ $\mathrm{P}_{1} \mathrm{~V}_{1}^{\gamma}=\mathrm{P}_{2} \mathrm{~V}_{2}^{\gamma}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\left(\frac{8 \mathrm{~V}_{1}}{\mathrm{~V}_{1}}\right)^{5 / 3}=\left(2^{3}\right)^{5 / 3}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=2^{5}=32$ $\mathrm{P}_{2}=32 \mathrm{P}_{1}$
J and K CET- 2000
Kinetic Theory of Gases
138940
An ideal gas undergoes a process that has the net effect of doubling its temperature and doubling its pressure. If $V$ was the initial volume of the gas, the final volume is
1 $\mathrm{V}$
2 $2 \mathrm{~V}$
3 $4 \mathrm{~V}$
4 $\frac{\mathrm{V}}{4}$
Explanation:
A For an ideal gas equation, $\quad \mathrm{PV}=\mathrm{nRT}$ $\therefore \quad \frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{P}_{2} \mathrm{~V}_{2}}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{2 \mathrm{P}_{1} \mathrm{~V}_{2}}=\frac{\mathrm{T}_{1}}{2 \mathrm{~T}_{1}} \quad\left[\because \mathrm{P}_{2}=2 \mathrm{P}_{1}, \mathrm{~T}_{2}=2 \mathrm{~T}_{1}\right]$ $\frac{\mathrm{V}_{1}}{2 \mathrm{~V}_{2}}=\frac{\mathrm{T}_{1}}{2 \mathrm{~T}_{1}}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{1}{1} \Rightarrow \mathrm{V}_{2}=\mathrm{V}_{1}=\mathrm{V} \quad\left(\because \mathrm{V}_{1}=\mathrm{V}\right)$
J and K CET- 1997
Kinetic Theory of Gases
138943
The volume of a gas at $20^{\circ} \mathrm{C}$ is $100 \mathrm{~cm}^{3}$ at 1 atmospheric pressure. If it is heated to $100^{\circ} \mathrm{C}$, its volume becomes $125 \mathrm{~cm}^{3}$ at the same pressure, then coefficient of gas at constant pressure is
1 $3.6 \times 10^{-3} /{ }^{\circ} \mathrm{C}$
2 $3.6 \times 10^{-4} /{ }^{\circ} \mathrm{C}$
3 $2.5 \times 10^{-3} /{ }^{\circ} \mathrm{C}$
4 $1.25 \times 10^{-3} /{ }^{\circ} \mathrm{C}$
Explanation:
A Given, Initial volume of a gas $\left(\mathrm{V}_{\mathrm{i}}\right)=100 \mathrm{~cm}^{3}$ Initial temperature of a gas $\left(\mathrm{T}_{\mathrm{i}}\right)=20^{\circ} \mathrm{C}$ Final volume of a gas $\left(\mathrm{V}_{\mathrm{f}}\right)=125 \mathrm{~cm}^{3}$ Final temperature of gas $\left(\mathrm{T}_{\mathrm{f}}\right)=100^{\circ} \mathrm{C}$ We know that, $\mathrm{V}_{\mathrm{f}}=\mathrm{V}_{\mathrm{i}}(1+\gamma \Delta \mathrm{T})$ $125=100[1+\gamma \times(100-20)]$ $1+\gamma(80)=\frac{125}{100}$ $1+80 \times \gamma=\frac{5}{4}$ $80 \times \gamma=\frac{5}{4}-1$ $80 \times \gamma=\frac{1}{4}$ $\gamma =\frac{1}{80 \times 4}$ $\gamma =3.125 \times 10^{-3} /{ }^{\circ} \mathrm{C}$ $\gamma \approx 3.6 \times 10^{-3} /{ }^{\circ} \mathrm{C}$
UP CPMT-2001
Kinetic Theory of Gases
138945
1 mole of an ideal gas at an initial temperature of $T K$ does $6 R$ joule of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is $5 / 3$, then final temperature of the gas will be
1 $(\mathrm{T}-4) \mathrm{K}$
2 $(\mathrm{T}+4) \mathrm{K}$
3 $(\mathrm{T}-2.4) \mathrm{K}$
4 $(\mathrm{T}+2.4) \mathrm{K}$
Explanation:
A Given, Initial temperature $=\mathrm{TK}$ Adiabatic work $=6 \mathrm{R}$ Joule $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma=5 / 3$ We know that, $\text { Work done }=\frac{\mathrm{P}_{1} \mathrm{~V}_{1}-\mathrm{P}_{2} \mathrm{~V}_{2}}{\gamma-1}$ $=\frac{\mathrm{nRT}_{1}-\mathrm{nRT}_{2}}{(\gamma-1)}$ $[\because \mathrm{PV}=\mathrm{nRT} \text{ Ideal gas equation}]$ $6 \mathrm{R}=\frac{\mathrm{nR}\left[\mathrm{T}_{1}-\mathrm{T}_{2}\right]}{\gamma-1}$ $6=\frac{\mathrm{T}-\mathrm{T}_{2}}{5 / 3-1}$ $6=\frac{3}{2}\left[\mathrm{~T}-\mathrm{T}_{2}\right]$ $4=\mathrm{T}-\mathrm{T}_{2}$ $\therefore \quad \mathrm{T}_{2}=(\mathrm{T}-4) \mathrm{K}$
138938
A monoatomic gas $\left(\gamma=\frac{5}{3}\right)$ is suddenly compressed to $1 / 8^{\text {th }}$ of the original volume adiabatically, then the pressure of the gas will become how many times larger than pressure originally?
1 32
2 $40 / 3$
3 8
4 same as before
Explanation:
A Given that, For monoatomic gas, $\gamma=\frac{5}{3}$ As we know that, for adiabatic process $\mathrm{PV}^{\gamma}=\text { constant }$ $\mathrm{P}_{1} \mathrm{~V}_{1}^{\gamma}=\mathrm{P}_{2} \mathrm{~V}_{2}^{\gamma}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\left(\frac{8 \mathrm{~V}_{1}}{\mathrm{~V}_{1}}\right)^{5 / 3}=\left(2^{3}\right)^{5 / 3}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=2^{5}=32$ $\mathrm{P}_{2}=32 \mathrm{P}_{1}$
J and K CET- 2000
Kinetic Theory of Gases
138940
An ideal gas undergoes a process that has the net effect of doubling its temperature and doubling its pressure. If $V$ was the initial volume of the gas, the final volume is
1 $\mathrm{V}$
2 $2 \mathrm{~V}$
3 $4 \mathrm{~V}$
4 $\frac{\mathrm{V}}{4}$
Explanation:
A For an ideal gas equation, $\quad \mathrm{PV}=\mathrm{nRT}$ $\therefore \quad \frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{P}_{2} \mathrm{~V}_{2}}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{2 \mathrm{P}_{1} \mathrm{~V}_{2}}=\frac{\mathrm{T}_{1}}{2 \mathrm{~T}_{1}} \quad\left[\because \mathrm{P}_{2}=2 \mathrm{P}_{1}, \mathrm{~T}_{2}=2 \mathrm{~T}_{1}\right]$ $\frac{\mathrm{V}_{1}}{2 \mathrm{~V}_{2}}=\frac{\mathrm{T}_{1}}{2 \mathrm{~T}_{1}}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{1}{1} \Rightarrow \mathrm{V}_{2}=\mathrm{V}_{1}=\mathrm{V} \quad\left(\because \mathrm{V}_{1}=\mathrm{V}\right)$
J and K CET- 1997
Kinetic Theory of Gases
138943
The volume of a gas at $20^{\circ} \mathrm{C}$ is $100 \mathrm{~cm}^{3}$ at 1 atmospheric pressure. If it is heated to $100^{\circ} \mathrm{C}$, its volume becomes $125 \mathrm{~cm}^{3}$ at the same pressure, then coefficient of gas at constant pressure is
1 $3.6 \times 10^{-3} /{ }^{\circ} \mathrm{C}$
2 $3.6 \times 10^{-4} /{ }^{\circ} \mathrm{C}$
3 $2.5 \times 10^{-3} /{ }^{\circ} \mathrm{C}$
4 $1.25 \times 10^{-3} /{ }^{\circ} \mathrm{C}$
Explanation:
A Given, Initial volume of a gas $\left(\mathrm{V}_{\mathrm{i}}\right)=100 \mathrm{~cm}^{3}$ Initial temperature of a gas $\left(\mathrm{T}_{\mathrm{i}}\right)=20^{\circ} \mathrm{C}$ Final volume of a gas $\left(\mathrm{V}_{\mathrm{f}}\right)=125 \mathrm{~cm}^{3}$ Final temperature of gas $\left(\mathrm{T}_{\mathrm{f}}\right)=100^{\circ} \mathrm{C}$ We know that, $\mathrm{V}_{\mathrm{f}}=\mathrm{V}_{\mathrm{i}}(1+\gamma \Delta \mathrm{T})$ $125=100[1+\gamma \times(100-20)]$ $1+\gamma(80)=\frac{125}{100}$ $1+80 \times \gamma=\frac{5}{4}$ $80 \times \gamma=\frac{5}{4}-1$ $80 \times \gamma=\frac{1}{4}$ $\gamma =\frac{1}{80 \times 4}$ $\gamma =3.125 \times 10^{-3} /{ }^{\circ} \mathrm{C}$ $\gamma \approx 3.6 \times 10^{-3} /{ }^{\circ} \mathrm{C}$
UP CPMT-2001
Kinetic Theory of Gases
138945
1 mole of an ideal gas at an initial temperature of $T K$ does $6 R$ joule of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is $5 / 3$, then final temperature of the gas will be
1 $(\mathrm{T}-4) \mathrm{K}$
2 $(\mathrm{T}+4) \mathrm{K}$
3 $(\mathrm{T}-2.4) \mathrm{K}$
4 $(\mathrm{T}+2.4) \mathrm{K}$
Explanation:
A Given, Initial temperature $=\mathrm{TK}$ Adiabatic work $=6 \mathrm{R}$ Joule $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma=5 / 3$ We know that, $\text { Work done }=\frac{\mathrm{P}_{1} \mathrm{~V}_{1}-\mathrm{P}_{2} \mathrm{~V}_{2}}{\gamma-1}$ $=\frac{\mathrm{nRT}_{1}-\mathrm{nRT}_{2}}{(\gamma-1)}$ $[\because \mathrm{PV}=\mathrm{nRT} \text{ Ideal gas equation}]$ $6 \mathrm{R}=\frac{\mathrm{nR}\left[\mathrm{T}_{1}-\mathrm{T}_{2}\right]}{\gamma-1}$ $6=\frac{\mathrm{T}-\mathrm{T}_{2}}{5 / 3-1}$ $6=\frac{3}{2}\left[\mathrm{~T}-\mathrm{T}_{2}\right]$ $4=\mathrm{T}-\mathrm{T}_{2}$ $\therefore \quad \mathrm{T}_{2}=(\mathrm{T}-4) \mathrm{K}$
