(B) For a monoatomic gas,the degrees of freedom $f = 3$.The molar specific heat at constant volume is given by ${C_V} = \frac{f}{2}R = \frac{3}{2}R$.U

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(B) For a monoatomic gas,the degrees of freedom $f = 3$.The molar specific heat at constant volume is given by ${C_V} = \frac{f}{2}R = \frac{3}{2}R$.Using Mayer's relation,${C_P} - {C_V} = R$,we can find the molar specific heat at constant pressure ${C_P}$.${C_P} = R + {C_V} = R + \frac{3}{2}R = \frac{5}{2}R$.

Class 11 Physics Kinetic Theory Molar Specific Heat of gas and relation between them (Mayer's formula)

Medium

The molar specific heat at constant pressure for a monoatomic gas is

A
$\frac{3}{2}R$
B
$\frac{5}{2}R$
C
$\frac{7}{2}R$
D
$4R$

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Class 11 Questions

Class 11

Physics Questions

Chapter

Kinetic Theory

Subtopic

Molar Specific Heat of gas and relation between them (Mayer's formula)

The molar specific heat at constant volume,${C_V}$,for a monoatomic gas is:

Medium

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Derive the ratio of $\frac{C_{P}}{C_{V}}$ for a diatomic gas.

Medium

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The ratio of the adiabatic to isothermal elasticities of a triatomic gas is

Easy

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Match the $\frac{C_{P}}{C_{v}}$ ratio for ideal gases with different types of molecules:

Molecule type $\frac{C_{P}}{C_{v}}$

$A$. Monoatomic $I$. $\frac{7}{5}$

$B$. Diatomic rigid molecules $II$. $\frac{9}{7}$

$C$. Diatomic non-rigid molecules $III$. $\frac{4}{3}$

$D$. Triatomic rigid molecules $IV$. $\frac{5}{3}$

Molecule type	$\frac{C_{P}}{C_{v}}$
$A$. Monoatomic	$I$. $\frac{7}{5}$
$B$. Diatomic rigid molecules	$II$. $\frac{9}{7}$
$C$. Diatomic non-rigid molecules	$III$. $\frac{4}{3}$
$D$. Triatomic rigid molecules	$IV$. $\frac{5}{3}$