Question

A system composed of 3 kg of this substance expands from an initial pressure of 400 kPa and a volume of 0.2 $m^2$ to a final pressure 100 kPa in a process in which process and volume are related by $P{V}^2=constant$.

If the expansion is quasi-static, then the magnitude of q is ______.

• A. 80 kJ
• B. 60 kJ
• C. 40 kJ
• D. 120 kJ

Answer 1

Answer: (A)

80 kJ

The internal energy of a certain substance is given by the following equation:

$U=3PV+84$

Where U is given in kJ/kg, P is in kPa, and V is in $m^3/kg$.

$\Delta U=U_2-U_1=3(P_2{V}_2-P_1{V}_1)$

$\therefore \Delta U=3(P_2{V}_2-P_1{V}_2)$

Now $P_1{V}_1^2=P_2{V}_2^2$

$\therefore V_2=V_1{\left(\frac{P_1}{P_2}\right)}^{1/2}=0.2{\left(\frac{4}{1}\right)}^{1/2}=0.4$

$\Delta U=300(1\times 0.4-4\times 0.2)kJ=-120kJ$

For a quasi-static process

$W=\int PdV=\frac{P_2{V}_2-P_1{V}_1}{n-1}$

$=\frac{(1\times 0.4-4\times 0.2)100}{2-1}=-40kJ$

$\therefore \Delta U=q+W$

$q=-120+40=-80kJ$

$q$ is negative means the reaction is exothermic.

Hence, the correct option is $\left(A\right)$