Question

A cylinder contains one liter of air at room temperature ( $300\mathrm{K}$) and atmospheric pressure $\left({10}^5\mathrm{N}/{\mathrm{m}}^2\right)$. At one end of the cylinder is a massless piston, whose surface area is $0.01{\mathrm{m}}^2$. Suppose that you push the piston in very suddenly, exerting a force of $2000\mathrm{N}$. The piston moves only one millimeter, before it is stopped by an immovable barrier of some sort.

(a) How much work have you done on this system?

(b) How much heat has been added to the gas?

(c) Assuming that all the energy added goes into the gas (not the piston or cylinder walls), by how much does the internal energy of the gas increase?

(d) Use the thermodynamic identity to calculate the change in the entropy of the gas (once it has again reached equilibrium).

Short answer

(a) $W=2J$

(b) There is no exchange of heat.

(c) $dU=2J$

(d)$0.01{\mathrm{JK}}^{-1}$

Step-by-step solution

Part (a) Step 1 : Explanation of Solution

Given:

Force, $F=2000\mathrm{N}$

The piston moves a distance, $dx=1\mathrm{mm}={10}^{-3}\mathrm{m}$

Formula used:

The work done is,$W=Fdx$

Part (a) Step 2 : Calculation

The work done in moving the piston $1\mathrm{mm}$before it is stopped by an immovable barrier is

$$\begin{gathered} W=Fdx \\ =2000\mathrm{N}\times {10}^{-3}\mathrm{m} \\ =2\mathrm{J} \end{gathered}$$

Part (a) Step 3 : Conclusion

There is a work done of $2\mathrm{J}$in the system.

Part (b) Step 1 : Introduction

As the piston is hit suddenly, the air is compressed very quickly. This is a case of extreme adiabatic compression, in which the compression occurs so quickly that there is no gain or lost by the gas.

The massless piston is suddenly pushed and moved $1\mathrm{mm}$before it is stopped by an immovable barrier.

As the piston is suddenly moved so one can conveniently approximate the process as adiabatic.So, no heat is added to the system.

Part (b) Step 2 : Conclusion

There is no exchange of heat.

Part (c) Step 1 : Explanation of Solution

Given:

Work done, $W=2\mathrm{J}$

Heat absorbed by the system, $Q=0$

Formula used:

The first law of thermodynamics,$dU=Q+W$

Part (c) Step 2 : Calculation

The internal energy of the system is

$$\begin{gathered} dU=Q+W \\ =0+2 \\ =2\mathrm{J} \end{gathered}$$

Part (c) Step 3 : Conclusion

The internal energy of the system is $2\mathrm{J}$.

Part (d) Step 1 : Explanation of Solution

Given:

Internal energy, $dU=2J$

Pressure, $P={10}^5\mathrm{Pa}$

Temperature, $T=300\mathrm{K}$

Change in volume, $dV=A·dx=0.01\times 0.001={10}^{-5}{\mathrm{m}}^3$

Formula used:

The thermodynamic identity for an infinitesimal process is

$dU=TdS-PdV$

Part (d) Step 2 : Calculation

The change in entropy is,

$dS=\frac{dU+PUV}{T}$

$=\frac{23+{10}^5{P}_{a\times {10}^{-5}{\mathrm{m}}^3}}{300\mathrm{K}}$

$=0.01{\mathrm{JK}}^{-1}$

Part (d) Step 3 : Conclusion

The entropy change is $0.01{\mathrm{JK}}^{-1}$.