Heat and Thermodynamics 5 Question 52
53. At $27^{\circ} C$ two moles of an ideal monoatomic gas occupy a volume $V$. The gas expands adiabatically to a volume $2 V$.
Calculate
(1996, 5M)
(a) the final temperature of the gas,
(b) change in its internal energy,
(c) the work done by the gas during this process.
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Answer:
Correct Answer: 53. (a) $189 K$ (b) $-2767 J$ (c) $2767 J$
Solution:
- Given, $T _1=27^{\circ} C=300 K, V _1=V, V _2=2 V$
(a) Final temperature
In adiabatic process, $T V^{\gamma-1}=$ constant
$$ \therefore \quad T _1 V _1^{\gamma-1}=T _2 V _2^{\gamma-1} $$
or $\quad T _2=T _1\left(\frac{V _1}{V _2}\right)^{\gamma-1}=300\left(\frac{V}{2 V}\right)^{5 / 3-1}$
$$ \gamma=\frac{5}{3} \text { for a monoatomic gas } T _2 \approx 189 K $$
(b) Change in internal energy,
$$ \begin{aligned} \Delta U & =n C _V \Delta T \\ \Delta U & =(2)\left(\frac{3}{2} R\right)\left(T _2-T _1\right) \\ \Delta U & =2\left(\frac{3}{2}\right)(8.31)(189-300) J \\ \Delta U & =-2767 J \end{aligned} $$
(c) Work done
Process is adiabatic, therefore $\Delta Q=0$
and from first law of thermodynamics,
$$ \begin{aligned} \Delta Q & =\Delta W+\Delta U \\ \Delta W & =-\Delta U=-(-2767 J) \\ \Delta W & =2767 J \end{aligned} $$
