Answer to Question #123079 in Molecular Physics | Thermodynamics for vamsi

Question

The temperature of n moles of ideal gas is increased from T to 4T through a process for which pressure P = a T^(-1) ,where 'a' is a constant. The work done by the gas is ?

Answer 1

Change in temperature is ,$ΔT=4T−T=3T$

Given, gas obeys the law

$$P=aT^{-1}$$

Now, from ideal gas law, $PV=nRT$

and solving the above equation we get,

$$PV^{\frac{1}{2}}=constant$$

Thus, process is poly-tropic.

We, know that molar heat capacity is

$$C=C_v+\frac{R}{1-x}$$

​Hence,

$$C=C_v+2R$$

Now, we also know that, heat absorbed is

$$ΔQ=nCΔT=ΔU+ΔW=nC_v\Delta T+\Delta W$$

Therefore, we get, work done is

$$\Delta W=n\Delta T(C-C_v)=6nRT$$

Note: $R$ is universal gas constant.