Answer to Question #273339 in Molecular Physics | Thermodynamics for B H Khan

Question

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

u = 3.56 pv + 184

where u is given in kJ/kg, p is in kPa and v is in m3

/kg.

A system composed of 3 kg of this substance expands from an initial pressure of 500

kPa and volume of 0.22 m3

to final pressure 100 kPa in a process in which pressure and

volume are related by pv1.2 = Constant. If expansion is quasi-static, find Q, ∆U and W

for the process.

Answer 1

$u=3.56pv+184 \\ ∆u=u_2-u_1 \\ ∆u=3.56(p_2V_2-p_1V_1)\\ p_1V_1^{1.2}=p_2V_2^{1.2}\\ V_2=V_1(\frac{p_1}{p_2})^{\frac{1}{1.2}} \\ V_2=0.22\times(\frac{500}{100})^{\frac{1}{1.2}}=0.841 \; m^3 \\ ∆U= 3.56(100\times10^3\times 0.841-500\times10^3\times 0.22) \\ ∆U=3.56 (100\times10^3(0.841-5 \times 0.22 ) )\\ ∆U=-92.204 \; kJ$

Quasi static process

$W=\smallint p dV =\frac{p_1V_1-p_2V_2}{n-1} \\ W=\frac{500\times10^3\times 0.21-100 \times10^3 \times 0.841}{1.2-1}=104.5 \; J \\ Q=∆U+W=-92.204+104.5=12.296 \; kJ Q=12.3 \; kJ$