Question

Two moles of helium gas undergo a cyclic process as shown in figure. Assuming the gs to be ideal, calculate the following quantities in this process. The net change in the heat energy in $J$.

Answer 1

Since the process is cyclic, change in internal energy in one cycle is $0$.

Hence net change in heat energy is equal to the negative of net work done.

$W_{AB}=P(V_B-V_A)=2R(T_B-T_A)$

$W_{BC}=-2R{T}_{B}ln\left(\frac{P_B}{P_C}\right)=-2R\left(400\right)ln\left(2\right)$

$W_{CD}=P(V_D-V_C)=2R(T_D-T_C)=-W_{AB}$

$W_{DA}=-2R{T}_{D}ln\left(\frac{P_D}{P_A}\right)=2R\left(300\right)ln\left(2\right)$

${Q=-W}_{net}=2\left(8.314\right)ln\left(2\right)(400-300)\approx 1153J$