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# Solve it, One moleof anideal monatomic gas undergoes a process described by the equation PV3 = constant. The heat capacity of the gas during this process is

One mole of an ideal monatomic gas undergoes a process described by the equation PV3 = constant. The heat capacity of the gas during this process is

• Option 1)

• Option 2)

• Option 3)

2 R

• Option 4)

R

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As we learnt in

Specific heat capacity at constant pressure -

$C_{p}= C_{v}+R$

$= \left ( \frac{f}{2}+1 \right )R$

- wherein

f = degree of freedom

R= Universal gas constant

Given process is PV3= constant

Heat capacity $=\frac{dQ}{ndT}$

PV3 = K      or     $P=\frac{K}{V^{3}}$

Work done = $\int PdV=\int \frac{K}{V^{3}}.dV$

$\left.\begin{matrix} =\frac{K.V^{-3}}{-3+1} \end{matrix}\right|_{i}^{f}$

$=-\frac{K}{2}.\left(\frac{1}{V_{f}^{2}}-\frac{1}{V_{i}^{2}} \right )$

$=-\frac{1}{2}.\left(\frac{K}{V_{f}^{2}}-\frac{K}{V_{i}^{2}} \right )$

$PV=\frac{K}{V^{2}}$        or     $\frac{K}{V^{2}}=nRT$

Work = $-\frac{1}{2}\left(nRT_{f}-nRT_{i} \right )=-\frac{nR}{2}.\Delta T$

$\therefore \Delta Q=nC_{V}\Delta T+\left(\frac{-nRT \Delta T}{2} \right )$

For monoatomic ideal gas

$C_{V}=\frac{3R}{2}$

$\Rightarrow \Delta Q=n\frac{3R}{2}\Delta T-\frac{nR}{2}\Delta T=nR \Delta T$

$\therefore C=\frac{\Delta Q}{n\Delta T}=R$

Option 1)

Incorrect

Option 2)

Incorrect

Option 3)

2 R

Incorrect

Option 4)

R

Correct

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