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In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation VT = k, where k is a constant. In this process, the temperature of the gas is increased by ∆T. The amount of heat absorbed by gas is (where, R is gas constant)
1. \(\frac{1}{2}kR\Delta T\)
2. \({\rm{\;}}\frac{{2k}}{3}\Delta T\)
3. \(\frac{1}{2}R\Delta T\)
4. \(\frac{3}{2}R\Delta T\)

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Correct Answer - Option 3 : \(\frac{1}{2}R\Delta T\)

Concept:

The polytropic process can describe gas expansion and compression which include heat transfer. The exponent n is known as the polytropic index and it may take on any value from 0 to ∞, depending on the particular process.

\(\Delta W = \frac{{{p_2}{V_2} - {p_1}{V_1}}}{{1 - x}}\left( {{\rm{\;for\;}}x \ne 0} \right)\) 

Calculation:

Given, VT = k, (k is constant)

\({\rm{or\;T}} \propto \frac{1}{{\rm{V}}}\;\)     ----(1)

Using ideal gas equation,

pV = nRT

pV ∝ T

\( \Rightarrow {\rm{pV}} \propto \frac{1}{{\rm{V}}}\)

or pV2 = constant      ----(2)

i.e. a polytropic process with x = 2

(Polytropic process means, pVx = constant)

We know that, work done in a polytropic process is given by

\(\Delta W = \frac{{{p_2}{V_2} - {p_1}{V_1}}}{{1 - x}}\left( {{\rm{\;for\;}}x \ne 0} \right)\)      ----(3)

\({\rm{and,\;\Delta W}} = {\rm{pV\;in\;}}\left( {\frac{{{{\rm{V}}_2}}}{{{{\rm{V}}_1}}}} \right)\left( {{\rm{\;for\;x}} = 1} \right)\)

Here, x = 2

\(\therefore {\rm{\Delta W}} = \frac{{{{\rm{p}}_2}{{\rm{V}}_2} - {{\rm{p}}_1}{{\rm{V}}_1}}}{{1 - {\rm{x}}}} = \frac{{{\rm{nR}}\left( {{{\rm{T}}_2} - {{\rm{T}}_1}} \right)}}{{1 - {\rm{x}}}}\)

\(\Rightarrow {\rm{\Delta W}} = \frac{{{\rm{nR\Delta T}}}}{{1 - 2}} = - {\rm{nR\Delta T}}\)        ----(4)

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

\({\rm{\Delta U}} = \frac{3}{2}{\rm{R\Delta T}}\)     ----(5)

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

∆Q = ∆W + ∆U

\(= \frac{3}{2}{\rm{R\Delta T}} - {\rm{R\Delta T}}\)

\(\Rightarrow {\rm{\Delta Q}} = \frac{1}{2}{\rm{R\Delta T}}\)

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