Answer to Question #115780 in Mechanical Engineering for Smith

Question

Air initially at 100 0F and 100 psia and occupying a volume of 0.5 ft3 undergoes a reversible non-flow constant temperature process such that the final pressure becomes 20 psia. find the work done, ft.lbf.

Answer 1

The equation of an isothermal process with ideal gas (like air) is

$$pV=nRT=const$$

The work $W$ of a reversible isothermal process is given by

$$W=-\int_{V_1}^{V_2}pdV=-nRT\int_{V_1}^{V_2}\frac{dV}{V}=-nRT\ln{\frac{V_2}{V_1}},$$

where $V_1, V_2$ are the initial and the final volume of the air respectively.

Substitute in the first equation the initial parameters

$$nRT=100psia\cdot0.5ft^3=50ft.lbf$$

We can find the final volume $V_2$ from the first equation

$$V_2=\frac{nRT}{p_2},$$

$$V_2=\frac{50ft.lbf}{20psia}=2.5ft^3.$$

Substitute to the second equation to find the work done

$$W=-50\cdot\ln{\frac{2.5}{0.5}}=-80.5ft.lbf.$$