Air that initially occupies 0.44 m3 at a gauge pressure of 55 kPa is expanded isothermally to a pressure of 101.3 kPa and then cooled at constant pressure until it reaches its initial volume. Compute the work done by the air. (Gauge pressure is the difference between the actual pressure and atmospheric pressure.)

[tex]\mathbf{W = \int ^{V_f}_{V_i} pdV} \\ \\ \\ \mathbf{W = nRT \int ^{V_f}_{V_i} \dfrac{dV}{V} } \\ \\ \\ \mathbf{ W= nRt In \dfrac{V_f}{V_i}}[/tex]

From ideal gas equation;

PV = nRT;
V = nRT/P

For the initial volume and final volume, we can say:

[tex]\mathbf{\dfrac{V_f}{V_i }= \dfrac{P_i}{P_f}}[/tex]

Then,

[tex]\mathbf{W = P_i V_i In \dfrac{P_i}{P_f}}[/tex]

where;

P_i = initial pressure
V_i = initial pressure

[tex]\mathbf{P_i= }[/tex] 0.55 × 10⁵ Pa + 1.013 × 10⁵ Pa

[tex]\mathbf{P_i= }[/tex] 1.563 × 10⁵ Pa

[tex]\mathbf{W = 1.563 \times 10^5 (0.44 ) In (\dfrac{1.563 \times 10^5 }{1.013 \times 10^5})}[/tex]

W = 3.00 × 10⁴ J

When there is constant pressure during the process, the work done by the gas is:

[tex]\mathbf{W = P_f(V_i - V_f)}[/tex]

where;

[tex]\mathbf{V_f = (\dfrac{P_i V_i}{P_f})}[/tex]

Thus;

[tex]\mathbf{W = P_f(V_i - \dfrac{P_i V_i}{P_f})}[/tex]

[tex]\mathbf{W = ( P_f -P_t)V_i}[/tex]

[tex]\mathbf{W = (1.013 \times 10^5 \ Pa - 1.563 \times 10^5 Pa)(0.44 \ m^3)}[/tex]

W = -2.42 × 10⁴ J.

Finally, the total work done by the air in the total process is:

W = 3.00 × 10⁴ J - 2.42 × 10⁴ J

W = 5.80 × 10³ J

Learn more about work done in an isothermal process here:

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