Question

A gas mixture contains one-mole $\mathop O\nolimits_2$ gas and one mole of $He$ gas. Find the ratio of specific heat at constant pressure to that at constant volume of gaseous mixture.
(A) 2
(B) 1.5
(C) 2.5
(D) 4

Answer 1

The Heat capacity is the change in temperature of a given sample for a given amount of heat. TheSI Units of specific heat is joule per kelvin $(J{\text{ }}k{g^{ - 1}}).$Specific Heat Capacity is different from Heat Capacity, which represents the amount of heat needed to raise the temperature of a 1kg mass of a substance ${1^o}C$. If the heat transfer to the sample is done when it is held at constant pressure, then the specific heat obtained using this method is called Molar Specific Heat Capacity at Constant Pressure. If the heat transfer to the sample is done when the volume of the sample is held constant, then the specific heat obtained using this method is called Molar Specific Heat Capacity at Constant Volume.
We Should know the ${C_P}and{\text{ }}{C_V}$ values of monoatomic gases and diatomic gases.
$\mathop O\nolimits_2$ Is a diatomic gas.
$He$ Is a monatomic gas.
$\dfrac{{\mathop C\nolimits_P }}{{\mathop C\nolimits_V }} = R$, We have to find R.

Complete step by step answer:
Method 1:
-$\mathop O\nolimits_2$ is a diatomic gas.
$\mathop C\nolimits_P = \dfrac{7}{2}R$
$\mathop C\nolimits_V = \dfrac{5}{2}R$
-$He$ is a monatomic gas.
$\mathop C\nolimits_P = \dfrac{5}{2}R$
$\mathop C\nolimits_V = \dfrac{3}{2}R$
So the $\mathop C\nolimits_P$ value of gaseous mixture is,
\mathop C\nolimits_P $$$$\left( {Mixture} \right){\text{ }} = \;$$$$\dfrac{{\mathop C\nolimits_P (\mathop O\nolimits_2 ) + \mathop C\nolimits_P (He)}}{2}

= \dfrac{{\dfrac{7}{2}R + \dfrac{5}{2}R}}{2} \\\ = \dfrac{{12}}{4}R \\\ \end{gathered} $$ $$\mathop C\nolimits_V $$ Value of gaseous mixture is, $$\mathop C\nolimits_V $$$$\left( {Mixture} \right){\text{ }} = \;$$$$\dfrac{{\mathop C\nolimits_V (\mathop O\nolimits_2 ) + \mathop C\nolimits_V (He)}}{2}$$ $$\begin{gathered} = \dfrac{{\dfrac{5}{2}R + \dfrac{3}{2}R}}{2} \\\ = \dfrac{8}{4}R \\\ \end{gathered} $$ So,$$\dfrac{{\mathop C\nolimits_P }}{{\mathop C\nolimits_V }} = R$$ $$\dfrac{{\mathop C\nolimits_P }}{{\mathop C\nolimits_V }} =\dfrac{{\dfrac{{12}}{4}R}}{{\dfrac{8}{4}R}}$$ $$\begin{gathered} = \dfrac{{12}}{8} = \dfrac{3}{2} \\\ = 1.5 \\\ \end{gathered} $$ The Option (B) is correct. **Method 2:** Helium $$\left( {He} \right)$$ is a monatomic gas. Oxygen $$\left( {O2} \right)$$ is a diatomic gas. For a monatomic gas and a diatomic gas, value of $$\mathop C\nolimits_V $$is $$\dfrac{3}{2}R$$ and $$\dfrac{5}{2}R$$ respectively. For a gaseous mixture, $$\;\mathop C\nolimits_V (mixture){\text{ }} = {\text{ }}\dfrac{{\left[ {\mathop n\nolimits_1 {\text{ }} \cdot \mathop C\nolimits_{\mathop V\nolimits_1 } {\text{ }} + {\text{ }}\mathop n\nolimits_2 \cdot \mathop C\nolimits_{\mathop V\nolimits_2 } } \right]}}{{\left[ {\mathop n\nolimits_1 + \mathop n\nolimits_2 } \right]}}{\text{ }}$$ Hence for the given mixture, $$\;\mathop C\nolimits_V (mixture){\text{ = }}\dfrac{{[2 \cdot \dfrac{3}{2} + 1 \cdot \dfrac{5}{2}]}}{{2 + 1}}{\text{R }}$$ $$\begin{gathered} = \dfrac{{\dfrac{{11}}{2}}}{3}R \\\ = \dfrac{{11}}{6}R \\\ \end{gathered} $$ Now since $$\begin{gathered} \;\mathop C\nolimits_P - \;\mathop C\nolimits_V = R \\\ \;\mathop C\nolimits_P = \mathop C\nolimits_V + R \\\ \mathop C\nolimits_p = \dfrac{{11}}{6}R + R \\\ \end{gathered} $$. $$ = \dfrac{{17}}{6}R$$ So we get, $$\dfrac{{\mathop C\nolimits_P }}{{\mathop C\nolimits_V }} = \dfrac{{\dfrac{{17}}{6}R}}{{\dfrac{{11}}{6}R}}$$ $$ = \dfrac{{17}}{{11}}$$ $$ = 1.5$$ So, the Option (B) is correct. **Note:** Don’t interchange n1 and n2 in$$\;\mathop C\nolimits_V (mixture){\text{ }} = {\text{ }}\dfrac{{\left[ {\mathop n\nolimits_1 {\text{ }} \cdot \mathop C\nolimits_{\mathop V\nolimits_1 } {\text{ }} + {\text{ }}\mathop n\nolimits_2 \cdot \mathop C\nolimits_{\mathop V\nolimits_2 } } \right]}}{{\left[ {\mathop n\nolimits_1 + \mathop n\nolimits_2 } \right]}}{\text{ }}$$. Find either $$\mathop C\nolimits_V $$or $$\mathop C\nolimits_P $$and substitute it in$$\mathop C\nolimits_P - \;\mathop C\nolimits_V = R$$.