91Ó°ÊÓ

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,$W=Fdx$

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

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

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

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.

There is no exchange of heat.

Given:

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

Heat absorbed by the system, $Q=0$

Formula used:

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

The internal energy of the system is

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

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

Given:

Internal energy, $dU=2J$

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

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

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

Formula used:

The thermodynamic identity for an infinitesimal process is

$dU=TdS-PdV$

The change in entropy is,

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

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

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

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