Heat and Thermodynamics 5 Question 19
20. $5.6 L$ of helium gas at STP is adiabatically compressed to $0.7 L$. Taking the initial temperature to be $T _1$, the work done in the process is
(2011)
(a) $\frac{9}{8} R T _1$
(b) $\frac{3}{2} R T _1$
(c) $\frac{15}{8} R T _1$
(d) $\frac{9}{2} R T _1$
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Answer:
Correct Answer: 20. (a)
Solution:
- At STP, $22.4 L$ of any gas is 1 mole.
$\therefore \quad 5.6 L=\frac{5.6}{22.4}=\frac{1}{4}$ moles $=n$
In adiabatic process, $T V^{\gamma-1}=$ constant
$\therefore \quad T _2 V _2^{\gamma-1}=T _1 V _1^{\gamma-1}$ or $T _2=T _1\left(\frac{V _1}{V _2}\right)^{\gamma-1}$ $\gamma=\frac{C _p}{C _V}=\frac{5}{3}$ for monoatomic He gas.
$\therefore \quad T _2=T _1\left(\frac{5.6}{0.7}\right)^{\frac{5}{3}-1}=4 T _1$
Further in adiabatic process, $Q=0$
$\therefore \quad W+\Delta U=0$
or $\quad W=-\Delta U=-n C _V \Delta T=-n\left(\frac{R}{\gamma-1}\right)\left(T _2-T _1\right)$
$$ =-\frac{1}{4}\left(\frac{R}{\frac{5}{3}-1}\right)\left(4 T _1-T _1\right)=-\frac{9}{8} R T _1 $$
