Answer
a) $V_{2}=0.0064792m^3$
b) $T_{2}=370.7^{\circ}C$
$P_{2}=21223.48kPa$
c)$\Delta U=1878,67kJ$
Work Step by Step
1) From table A-4:
$\upsilon_{1}=0.001157\frac{m^3}{kg}$
a) $V_{1}=m\upsilon_{1}=1.4kg*0.001157\frac{m^3}{kg}=0.0016198m^3$
$u_{1}=850.46\frac{kJ}{kg}$
2) $V_{2}=4V_{1}=4*0.0016198m^3=0.0064792m^3$
$\upsilon_{2}=\frac{V_{2}}{m}=\frac{0.0064792m^3}{1.4kg}=0.004628\frac{m^3}{kg}$
Interpolating from table A-4:
b) $T_{2}=370.7^{\circ}C$
$P_{2}=21223.48kPa$
$u_{2}=2192.37\frac{kJ}{kg}$
c)$\Delta U=m\Delta u=1.4kg*(2192.37\frac{kJ}{kg}-850.46\frac{kJ}{kg})=1878,67kJ$
