# 机械科学 Mechanical Science GENG0003W1-01

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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.

## GENG0003W1-01COURSE 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,
$$W_{1-2}=\frac{p_{1} \forall_{1}-p_{2} \forall_{2}}{\gamma-1}=\frac{(500 \times 0.2-100 \times 0.6313) 10^{3}}{(1.4-1) \times 10^{3}}=92.175 \mathrm{~kJ}$$