Answer to Question #130709 in Molecular Physics | Thermodynamics for kishan

Question

1 kg of gas is contained in a Piston-cylinder assembly at 10 bar. The gas is allowed to expand reversibly until volume becomes twice the initial value, by following PV2 = C. The gas is then cooled at constant pressure until the piston reaches to its original position. Finally heat is supplied to the fluid reversibly, with piston firmly locked, and due to which the fluid pressure rises to its initial value of 10 bar. If fluid has an initial volume of 0.05 m3. Determine the net work done.

Answer 1

The first process is isotermic, $T=const$. The work, that has done during the process:

$A_1=P_1V_1*ln\tfrac{V_2}{V_1}$. By the condition of the problem $V_2=2V_1$. Substitute it.

$A_1=P_1V_1*ln\tfrac{2V_1}{V_1}=P_1V_1*ln2$

The second process is isobaric $P=const$. The work:

$A_2=P_2(V_2-V_1)=P_2(2V_1-V_1)=P_2V_1$

$P_2$ we can find from the Boyle-Mariott law for the isotermic process:

$P_1V_1=P_2V_2$, $P_1V_1=2P_2V_1->P_1=2P_2->P_2=\frac{1}{2}P_2$

So

$A_2=\frac{1}{2}P_1V_1$

When piston has locked, it was the isochore process $V=const$. The work during this process:

$A_3=0$.

The total work:

$A=A_1+A_2+A_3$

$A=P_1V_1*ln2+\frac{1}{2}P_1V_1=P_1V_1(ln2+\frac{1}{2})$

Substitute numbers and take into account that $10 (bar)=1000000(Pa)$, $ln2=0.69$.

$A=1000000*0.05*(0.69+0.5)=50000*1.19=59500 (J)$

Answer: A = 59.5 (kJ).