Question

An ideal gas undergoes a process according to the relation
$P=(\frac{0.80}{V}+S)$ bar where volume is in $m^3$ . If the gas contracts from $1m^3$ to $0.7m^3$ and 9 kJ is rejected from the system, the change in enthalpy during the process is ____$kJ$.

• A. 18
• B. 19.534
• C. 20
• D. 16.534

Answer 1

The correct option is B 19.534

From TDs relation, we have

$TdS=dH-VdP$

$dH=TdS+VdP$

$dH=dQ+VdP$

$P=\frac{0.80}{V}+8$

$dP=-\frac{0.80}{V^2}dV$

$\Rightarrow dH=dQ+V(-\frac{0.80}{V^2})dV=dQ-\frac{0.80}{V}\times dV$

$=dQ-0.80\times {10}^2\left(\ln V{)}_{1.0}^{0.7}\right[1bar=100kPa]$

$=-9-0.80\times {10}^2\ln \left(\frac{0.7}{1}\right)$

$=-9+28.53=19.534kJ$