这是一份southampton南安普敦大学GENG0003W1-01作业代写的成功案例

If final pressure of the gas is $p_{3}$, for a constant volume process $3-1$,
$$
p_{3}=\frac{T_{3}}{T_{1}} p_{1}=\frac{323}{423} \times 10=7.6 \mathrm{bar}
$$
Let us find the mass of the gas $m$ and
$$
m=\frac{p_{1} \forall \forall_{1}}{R T_{1}}=\frac{10 \times 10^{5} \times 0.336}{293 \times 423}=2.7 \mathrm{~kg}
$$
Change in internal energy
$$
d U=U_{3}-U_{1}=m C_{7}\left(T_{3}-T_{1}\right)=2.7 \times 0.703(323-423)=-189.8 \mathrm{~kJ}
$$
The negative sign indicates that there is a decrease in internal energy.

GENG0003W1-01 COURSE NOTES ：

The path followed by the system, $p_{1} \forall_{1}^{\prime}=p_{2} \forall_{2}$.
So,
$$
\left(\frac{\forall_{2}}{\forall_{1}}\right)^{\gamma}=\frac{p_{1}}{p_{2}}
$$
and
$$
\forall_{2}=\left(\frac{p_{1}}{p_{2}}\right)^{1 / \gamma} \forall_{1}=\left(\frac{500}{100}\right)^{1 / L 4} \times 0.2=0.6313 \mathrm{~m}^{3}
$$
Hence work done,
$$
W_{1-2}=\frac{p_{1} \forall_{1}-p_{2} \forall_{2}}{\gamma-1}=\frac{(500 \times 0.2-100 \times 0.6313) 10^{3}}{(1.4-1) \times 10^{3}}=92.175 \mathrm{~kJ}
$$