Question

Consider the following PV diagram for a monoatomic gas. The ratio of work done by the gas to the change in internal energy of the gas will be ​

Answer 1

Answer:

3:1 is the correct answer.

Answer 2

The ratio of work done by the gas to the change in internal energy of the gas will be 3:1

Therefore, option (2) is correct.

Explanation:

At initial point from ideal gas equation

$PV=nRT$

or, $T=\frac{PV}{nR}$

At initial point

$T_0=\frac{P_0V_0}{nR}$

At final point

$T_f=\frac{2P_0\times 2V_0}{nR}$

$\implies T_f=\frac{4P_0V_0}{nR}$

Change in internal energy

$\Delta U=nC_v(T_f-T_0)$

$\implies \Delta U=nC_v(\frac{4P_0V_0}{nR}-\frac{P_0V_0}{nR})$

$\implies \Delta U=n\times\frac{3R}{2}(\frac{3P_0V_0}{nR})$

$\implies \Delta U=\frac{9P_0V_0}{2}$

Work done by the gas in expansion

$W=\frac{1}{2}(2V_0-V_0)\times (P_0+2P_0)$

$\implies W=\frac{3P_0V_0}{2}$

Ratio of work done to the change in internal energy

$\frac{W}{\Delta U}=\frac{9P_0V_0/2}{3P_0V_0/2}$

$\implies \frac{W}{\Delta U}=3$

$\implies W:\Delta U=3:1$

Hope this answer is helpful.

Know More:

Q: A mono atomic ideal gas sample is given heat Q. One half of this heat is used as work done by the gas and rest is used for increasing its internal energy. The equation of process in terms of volume and temperature is ?

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