Question

A horizontal piston/cylinder arrangement is placed in a constant-temperature bath. The piston sides in the cylinder with negligible friction and an external force holds it in place against an initial gas pressure of 14 bar. The initial gas volume is 0.03 m3. The external fore on the piston is reduced gradually and the gas expands isothermally as its volume triples. If the volume of the gas is related to its pressure so that the product PV is constant.
(a) What is the work by the gas in moving the external force?
(b) How much work would be done if the external force were suddenly reduced to half its initial value instead of being gradually
reduced?

Answer 1

Answer:

$W=46141.72~J$

Explanation:

Given:

initial pressure, $P_1=14~bar=1.4\times10^6~Pa$

initial volume, $V_1=0.03~m^3$

After isothermal expansion:

final volume, $V_2=3V_1=0.09~m^3$

a)

We have the work done in isothermal process as:

$W=P_1.V_1\ln(\frac{V_2}{V_1} )$

$W=1.4\times10^6\times 0.03\ln(\frac{0.09}{0.03} )$

$W=46141.72~J$ is the work by the gas in moving the external force

b)

When the force on piston is suddenly reduced to half the initial value then the process becomes near to adiabatic.

$P_1V_1^{\gamma}=P_2V_2^{\gamma}$

$14\times 0.03^{1.4}=P_2\times 0.09^{1.4}$

$P_2=3~bar$

Now, using the formula for work done in adiabatic process:

$W=\frac{P_1V_1-P_2V_2}{\gamma-1}$

$W=\frac{1.4\times10^6\times 0.03-3\times10^5\times 0.09}{1.4-1}$

$W=3750~J$