Courses
Courses for Kids
Free study material
Offline Centres
More
Store

For an ideal gas $\dfrac{{{{\text{C}}_{\text{p}}},{\text{m}}}}{{{{\text{C}}_{\text{v}}},{\text{m}}}} = \gamma$ . The molecular mass of the gas is M, its specific heat capacity at constant volume is:A.$\dfrac{{\text{R}}}{{{\text{M}}\left( {\gamma - 1} \right)}}$B.$\dfrac{{\text{M}}}{{{\text{R}}\left( {\gamma - 1} \right)}}$C.$\dfrac{{\gamma {\text{RM}}}}{{\gamma - 1}}$D.$\dfrac{{\gamma {\text{R}}}}{{{\text{M}}\left( {\gamma - 1} \right)}}$

Last updated date: 13th Jun 2024
Total views: 384.3k
Views today: 7.84k
Verified
384.3k+ views
Hint: Mayer’s formula tells the relation of heat capacities at constant pressure and volume with universal gas constant. Poisson ratio is given by $\dfrac{{{{\text{C}}_{\text{p}}}}}{{{{\text{C}}_{\text{v}}}}} = \gamma$ .

Complete step by step answer:
Molar heat capacity and specific heat capacity can be related as: ${{\text{C}}_{\text{m}}} = {\text{c}} \times {\text{M}}$ . According to Mayer’s formula; universal gas constant and heat capacities at constant pressure and volume can be related as: ${{\text{C}}_{\text{p}}} - {{\text{C}}_{\text{v}}} = {\text{R}}$ where ${{\text{C}}_{\text{p}}}{\text{ and }}{{\text{C}}_{\text{v}}}$ are molar heat capacity at constant pressure and constant volume respectively. Similarly, relation between universal gas constant and specific heat capacities at constant pressure and constant volume can be given as: ${{\text{c}}_{\text{p}}} - {{\text{c}}_{\text{v}}} = \dfrac{{\text{R}}}{{\text{M}}}$ where ${{\text{c}}_{\text{p}}}{\text{ and }}{{\text{c}}_{\text{v}}}$ are specific heat capacity at constant pressure and constant volume respectively and M is molecular weight.
According to Poisson’s ratio $\dfrac{{{{\text{C}}_{\text{p}}}}}{{{{\text{C}}_{\text{v}}}}} = \gamma$
where ${{\text{C}}_{\text{p}}}{\text{ and }}{{\text{C}}_{\text{v}}}$ molar heat capacity at constant pressure and constant volume respectively are.
Similarly, $\dfrac{{{{\text{c}}_{\text{p}}}}}{{{{\text{c}}_{\text{v}}}}} = \gamma$
where ${{\text{c}}_{\text{p}}}{\text{ and }}{{\text{c}}_{\text{v}}}$ are specific heat capacity at constant pressure and constant volume respectively.
Now, in order to find specific heat capacity at constant volume:
As we known, ${{\text{C}}_{\text{p}}} - {{\text{C}}_{\text{v}}} = {\text{R}}$ .
Dividing both the sides by ${{\text{C}}_{\text{v}}}$ ,
we get $\dfrac{{{{\text{C}}_{\text{p}}}}}{{{{\text{C}}_{\text{v}}}}} - \dfrac{{{{\text{C}}_{\text{v}}}}}{{{{\text{C}}_{\text{v}}}}} = \dfrac{{\text{R}}}{{{{\text{C}}_{\text{v}}}}}$ and as we know $\dfrac{{{{\text{C}}_{\text{p}}}}}{{{{\text{C}}_{\text{v}}}}} = \gamma$ ,
Therefore on solving we get $\gamma - 1 = \dfrac{{\text{R}}}{{{{\text{C}}_{\text{v}}}}}$ and as ${{\text{C}}_{\text{m}}} = {\text{c}} \times {\text{M}}$ , if we convert molar heat capacity to specific heat capacity, we get $\gamma - 1 = \dfrac{{\text{R}}}{{{{\text{c}}_{\text{v}}} \times {\text{M}}}}$
On rearranging we get: ${{\text{c}}_{\text{v}}} = \dfrac{{\text{R}}}{{{\text{M}}\left( {\gamma - 1} \right)}}$ .

Therefore, the correct option is A.

Note:
Heat capacity is the amount of heat required to raise the temperature by ${1^o}{\text{C or }}{1^o}{\text{K}}$ of a substance. It is represented by C. it can be given as: ${\text{C}} = \dfrac{{{\text{dQ}}}}{{{\text{dT}}}}$ . The unit of heat capacity is ${\text{J }}{{\text{K}}^{ - 1}}$ .
Molar heat capacity is the amount of heat required to raise the temperature by ${1^o}{\text{C or }}{1^o}{\text{K}}$ of 1 mole of a substance. It is represented by ${{\text{C}}_{\text{m}}}$ and it can be given as: ${{\text{C}}_{\text{m}}} = \dfrac{{\text{C}}}{\mu }$ . The unit of molar heat capacity is ${\text{J mo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}$ .
Specific heat capacity is defined as the amount of heat required to raise the temperature of 1g mass of the substance through ${1^o}{\text{C or }}{1^o}{\text{K}}$ . The SI unit of specific heat is ${\text{J k}}{{\text{g}}^{ - 1}}{{\text{K}}^{ - 1}}$ . It is given as ${\text{c}} = \dfrac{1}{{\text{m}}}\dfrac{{\Delta {\text{Q}}}}{{\Delta {\text{T}}}}$ .