Chemistry - Thermodynamics - I Question with Solution | TestHub

$T_A=300$ K.

$\frac{T_B}{T_A}=\frac{V_B}{V_A}=\frac{2V}{V}$

$T_B=2T_A=2\times 300=600$ K.

$T_C=600$ K.

$T_D=75$ K, and the volume at D will be:

$$T_A{V}_A^{\gamma -1}=T_D{V}_D^{\gamma -1}$$

$${\left(\frac{300}{75}\right)}^{1/(\gamma -1)}V=V_D$$

$$V_D=8V$$

The total work done $W$ is the sum of work done in each process:

$$W=W_{AB}+W_{BC}+W_{CD}+W_{DA}$$

$$W=[-1\times R\times (600-300)]+\left[-1\times R\times 600\ln \frac{8V}{2V}\right]+0+\left[1\times \frac{3}{2}R(300-75)\right]$$

$$W=-1605cal$$