Answer to Question #227655 in Molecular Physics | Thermodynamics for Unknown346307

Question

2. A cylinder contains 1 kg of a certain fluid at an initial pressure of 20 bar.

The fluid is allowed to expand reversibly behind a piston according to a law pV2 = constant until

the volume is doubled. The fluid is then cooled reversibly at constant pressure until the piston

regains its original position ; heat is then supplied reversibly with the piston firmly locked in

position until the pressure rises to the original value of 20 bar. Calculate the net work done by the

fluid, for an initial volume of 0.05 m3

Answer 1

For the first step, since the state law is $PV^2=k$, we can consider that $P=\frac{k}{V^2} \text{ and }P_1V_1^2=k=P_2V_2^2$ both satisfy that equation, thus we calculate the work for these conditions as:

$P_1=20\,bar=2\times10^6\,Pa \\ V_1=0.05\,m^3 \\ P_2=\frac{P_1V_1^2}{V_2^2}=\frac{P_1}{4}=5\,bar=5\times10^5\,Pa \\ V_2=2V_1=0.10\,m^3 \\ W_{1st\,step}=-\int p\,dV=-\int_{V_1}^{V_2} k\frac{dV}{V^2} \\ W_{1st\,step}=k[{\cfrac{1}{V}}]_{V_1}^{V_2}=P_1V_1^2[\frac{1}{V_2}-\frac{1}{V_1}] \\ W_{1st\,step}= P_1V_1[\frac{V_1}{V_2}-1] \\ W_{1st\,step}=(2\times10^6\,Pa)(0.05\,m^3)(\frac{0.05\,m^3}{0.10\,m^3}-1) \\ W_{1st\,step}=(2\times10^6)(0.05)(-\frac{1}{2})\,Pa\cdot m^3 (\frac{1\,kJ}{1000\,Pa\cdot m^3}) \\ W_{1st\,step}=-50\,kJ$

For the second step, we have to consider that the process happens with constant pressure, thus the work can be found as:

$P_1=P_2=5\,bar=5\times10^5\,Pa \\ V_1=0.10\,m^3 \\ V_2=\frac{V_1}{2}=0.05\,m^3 \\ W_{2nd\,step}=-\int p\,dV=-P_1\int_{V_1}^{V_2} dV \\ W_{2nd\,step}=-P_1[V]_{V_1}^{V_2}=-P_1(V_2-V_1) \\ W_{2nd\,step}=-(5\times10^5\,Pa)(0.05-0.10)m^3 \\ W_{2nd\,step}=-(5\times10^5)(-0.05)\,Pa\cdot m^3 (\frac{1\,kJ}{1000\,Pa\cdot m^3}) \\ W_{2nd\,step}=25\,kJ$

For the third step, since the process occurs at constant volume (because the piston is locked in its position) there is no work done thus:

$P_1=5\,bar=5\times10^5\,Pa \\ P_2=20\,bar=2\times10^6\,Pa \\ V_2=V_1=0.05\,m^3 \\ W_{3rd\,step}=-\int p\,dV \\ \text{ (since dV=0) } W_{3rd\,step}= 0 \,kJ$

The total work done is $W_T=W_{1st\,step}+W_{2nd\,step}+W_{3rd\,step}=(-50+25+0)kJ=-25\,kJ.$

Reference:

• Sears, F. W., & Zemansky, M. W. (1973). University physics.