Answer to Question #290847 in Molecular Physics | Thermodynamics for eysiii

Question

3. A working substance undergoes a reversible non flow process during which P

= (-3V + 100) psia, where V changes from 10 ft3/min to 30 ft3/min. The change

of internal energy for the process is – 100 BTU/min. For a non-flow process,

determine a) work (BTU/min) b) heat transferred (BTU/min).

Answer 1

$P=(-3V+100)psia$

a) Work$=\int_{V_1}^{V_2}PdV$

$=\int_{10}^{30}-3V+100dV$

$\frac{-3V^2}{2}+100V\mid_{10}^{30}$

$\frac{-3(30^2)}{2}+100(30)+\frac{3(10^2)}{2}-100(10)\\~~~~~~=800BTU/min$

b) Heat transferred$= 100-800= -700BTU/min$

$Note: E_{int}=Q-W$