Heat and Thermodynamics 5 Question 8
8. In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation $V T=k$, where $k$ is a constant. In this process, the temperature of the gas is increased by $\Delta T$. The amount of heat absorbed by gas is (where, $R$ is gas constant)
(2019 Main, 11 Jan II)
(a) $\frac{1}{2} k R \Delta T$
(b) $\frac{2 k}{3} \Delta T$
(c) $\frac{1}{2} R \Delta T$
(d) $\frac{3}{2} R \Delta T$
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Answer:
Correct Answer: 8. (c)
Solution:
- Given, $V T=k,(k$ is constant )
or
$$ T \propto \frac{1}{V} \cdots(i) $$
Using ideal gas equation,
$$ \begin{aligned} p V & =n R T \\ p V & \propto T \Rightarrow p V \propto \frac{1}{V} \\ p V^{2} & =\text { constant } \cdots(ii) \end{aligned} $$
or
i.e a polytropic process with $x=2$.
(Polytropic process means, $p V^{x}=$ constant)
We know that, work done in a polytropic process is given by
$$ \Delta W=\frac{p _2 V _2-p _1 V _1}{1-x}(\text { for } x \neq 1) \cdots(iii) $$
and, $\quad \Delta W=p V \ln \left(\frac{V _2}{V _1}\right)($ for $x=1)$
Here, $x=2$,
$$ \begin{array}{ll} \therefore & \Delta W=\frac{p _2 V _2-p _1 V _1}{1-x}=\frac{n R\left(T _2-T _1\right)}{1-x} \\ \Rightarrow & \Delta W=\frac{n R \Delta T}{1-` 2}=-n R \Delta T \cdots(iv) \end{array} $$
Now, for monoatomic gas change in internal energy is given by
$$ \Delta U=\frac{3}{2} R \Delta T \cdots(v) $$
Using first law of thermodynamics, heat absorbed by one mole gas is
$\Delta Q=\Delta W+\Delta U=\frac{3}{2} R \Delta T-R \Delta T \Rightarrow \Delta Q=\frac{1}{2} R \Delta T$
