Answer to Question #279578 in Mechanics | Relativity for jean

Question

one pound of air at 15 psia, 300 f is heated at constant volume reversibly until its temperature becomes 600 F. Find P2,W,

Answer 1

Given quantities:

$P_1=15 psia$

$T_1=300F = (300-32)*\frac{5}{9}+273=422K$

$T_1=600F = (600-32)*\frac{5}{9}+273=588K$

$\frac{P}{T}=const$

$\frac{P_1}{T_1}=\frac{P2}{T_2} \to P_2=\frac{T_2}{T_1}P_1=\frac{588}{422}*15=20.9 psia$