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A cylinder consists of a gas mixture of helium and oxygen. If the mass of helium is $4 \mathrm{~g}$ and the mass of oxygen is $32 \mathrm{~g}$, then the ratio of specific heat of the mixture $\left(C_p / C_V\right)$
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The correct answer is:
3/2
Mass of helium, $m_{\mathrm{He}}=4 \mathrm{~g}$
Mass of oxygen, $m_{\mathrm{O}}=32 \mathrm{~g}$
Helium is a monoatomic gas.
$\therefore$ Degrees of freedom, $f_1=3$
Oxygen is a diatomic gas.
$\therefore$ Degrees of freedom, $f_2=5$
$4 \mathrm{~g}$ helium contains $N_1=N_A=6.023 \times 10^{23}$ number of molecules
$32 \mathrm{~g}$ oxygen contains $N_2=N_A$ number of molecules.
(As atomic mass of oxygen is $16 \mathrm{~g}$ )
$$
\begin{aligned}
\therefore \quad \frac{C_p}{C_V} & =\gamma \\
& =\frac{N_1\left(2+f_1\right)+N_2\left(2+f_2\right)}{N_1 f_1+N_2 f_2} \\
& =\frac{N_A(2+3)+N_A(2+5)}{3 N_A+5 N_A} \\
& =\frac{5+7}{3+5}=\frac{12}{8}=\frac{3}{2}
\end{aligned}
$$
Mass of oxygen, $m_{\mathrm{O}}=32 \mathrm{~g}$
Helium is a monoatomic gas.
$\therefore$ Degrees of freedom, $f_1=3$
Oxygen is a diatomic gas.
$\therefore$ Degrees of freedom, $f_2=5$
$4 \mathrm{~g}$ helium contains $N_1=N_A=6.023 \times 10^{23}$ number of molecules
$32 \mathrm{~g}$ oxygen contains $N_2=N_A$ number of molecules.
(As atomic mass of oxygen is $16 \mathrm{~g}$ )
$$
\begin{aligned}
\therefore \quad \frac{C_p}{C_V} & =\gamma \\
& =\frac{N_1\left(2+f_1\right)+N_2\left(2+f_2\right)}{N_1 f_1+N_2 f_2} \\
& =\frac{N_A(2+3)+N_A(2+5)}{3 N_A+5 N_A} \\
& =\frac{5+7}{3+5}=\frac{12}{8}=\frac{3}{2}
\end{aligned}
$$
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