Question

Two moles of helium gas undergoes a cyclic process as shown in the figure. Assume the gas to be ideal, then the net change in heat energy is:

• A. $-1153.53J$
• B. $+1153.53J$
• C. $+1553.53J$
• D. None of the above

Answer 1

The correct option is A $-1153.53J$

Work done in the process A to B is $P(V_2-V_1)=nR(T_2-T_1)=2\times 8.314\times 100=1662.8J$

Work done in the process B to C is $-nRTln\left(\frac{P_2}{P_1}\right)=2\times 8.314\times 400\times ln2=4610.26J$

Work done in the process C to D is $P(V_2-V_1)=nR(T_2-T_1)=-2\times 8.314\times 100=-1662.8J$

Work done in the process D to A is $-nRTln\left(\frac{P_2}{P_1}\right)=-2\times 8.314\times 300\times ln2=-3457.69J$

Total work done is $\approx 1153J$

From 1st law,

$\Delta U=q+w$

$\Delta U=0$

Hence, $q=-w=-1153J$

Hence, option A is correct.