Question

A piston cylinder device contains $0.05m^3$ of gas initially at 180 kPa. At this state, a linear spring that has a spring constant of $150kN/m$ is just touching the piston but exerting no force on it. Now heat is transferred to the gas, causing the piston to rise and to compress the spring until the volume inside the cylinder doubles. If the cross-sectional area of the piston is $0.25m^2$, Then the total work done $\left(inkJ\right)$ by the gas will be

• A. 12
• B. 14
• C. 15
• D. 18

Answer 1

The correct option is A 12

Assumption : The expansion process is quasi-equilibrium and spring is linear in the range of interest,

$V_2=2V_1=2\times 0.05=0.1m^3$

Displacement of the piston (and of the spring)

$x=\frac{\Delta V}{A}=\frac{0.1-0.05}{0.25}$

$=0.2m$

Additional pressue applied by the spring on the gas at final state

$p=\frac{F}{A}=\frac{kx}{A}=\frac{150\times 0.2}{0.25}=120kPa$

Without the spring, the pressure of the gas would remain constant at 180 kPa while the piston is rising. But under the effect of the spring, The pressure rises linearly from 180 kPa to 180+120=300 kPa at the final state

Total work done=work done against the atmosphere +work done against the spring

$=\text{Area I}+\text{Area II}$

$=\left(\frac{300+180}{2}\right)(0.1-0.05)=12kJ$

Alternate :
Total work done by the gas = Work done in expansion + Work saved in the form of potential energy of spring

$=180(0.1-0.05)+\frac{1}{2}\times 150\times {0.2}^2=12kJ$