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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)

    \frac{3} {2}R

  • Option 2)

    \frac{5} {2}R

  • Option 3)

    2 R

  • Option 4)

    R

 
Answers (1)
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As we learnt in

Specific heat capacity at constant pressure -

C_{p}= C_{v}+R

= left ( frac{f}{2}+1 
ight )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)

\frac{3} {2}R

Incorrect

Option 2)

\frac{5} {2}R

Incorrect

Option 3)

2 R

Incorrect

Option 4)

R

Correct

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