Answer to Question #236950 in Molecular Physics | Thermodynamics for john

Question

Half a pound of gas undergoes a reversible non-flow process with the pressure remaining constant at 50 psia while the volume increases from 4 ft' to 8ft. Find the work done, Btu/lb

Answer 1

We only have to substitute the solution for the integral of the work when the pressure remains constant:

$W=\int_{V_1}^{V_2}pdV =p{\Delta V}=p(V_2-V_1) \\ \text{ } \\ \text{then, we substitute and find the value for W:} \\W=\dfrac{(50\,psia)(8-4)\,ft^3}{0.5 \,lb} \\ \text{ } \\ \text{finally, we use the relation between the variables to find the value for W in Btu/lb:} \\W=400\,\dfrac{psia\cdot ft^3}{lb} \times\cfrac{1\,Btu}{5.40362\,psia\cdot ft^3} \\W=74.0244\, \cfrac{Btu}{lb}$

In conclusion, we find that the work done is equal to W=74.0244 But/lb.

Reference:

• Sears, F. W., & Zemansky, M. W. (1973). University physics.