Heat and Thermodynamics 1 Question 4

4. 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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Solution:

  1. Given, $V T=k$, $(k$ is constant $)$

$$ \text { or } \quad T \propto \frac{1}{V} $$

Using ideal gas equation,

$$ \begin{array}{ll} & p V=n R T \\ & p V \propto T \\ \Rightarrow \quad & p V \propto \frac{1}{V} \\ \text { or } \quad & p V^{2}=\text { constant } \end{array} $$

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

$$ \Delta W=\frac{p _2 V _2-p _1 V _1}{1-x}(\text { for } x \neq 1) $$

and, $\quad \Delta W=p V \ln \left(\frac{V _2}{V _1}\right)($ for $x=1)$

Here, $x=2$,

$$ \begin{array}{ll} \therefore & \Delta W=\frac{p _2 V _2-p _1 V _1}{1-x}=\frac{n R\left(T _2-T _1\right)}{1-x} \\ \Rightarrow & \Delta W=\frac{n R \Delta T}{1-` 2}=-n R \Delta T \end{array} $$

Now, for monoatomic gas change in internal energy is given by

$$ \Delta U=\frac{3}{2} R \Delta T $$

Using first law of thermodynamics, heat absorbed by one mole gas is

$$ \begin{aligned} \Delta Q & =\Delta W+\Delta U=\frac{3}{2} R \Delta T-R \Delta T \\ \Rightarrow \quad \Delta Q & =\frac{1}{2} R \Delta T \end{aligned} $$



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