Kinetic Theory of Gases MCQs for NEET — Physics Questions with Answers

4. According to the equilibrium theorem, the molar heat capacities should be independent of

    temperature. However, variations in ${\mathrm{C}}_{\mathrm{V}}$ and ${\mathrm{C}}_{\mathrm{P}}$ are observed as the temperature changes. At very

    high temperatures, vibrations are also important and that affects the values of ${\mathrm{C}}_{\mathrm{V}}$ and ${\mathrm{C}}_{\mathrm{P}}$ for

    diatomic and polyatomic gases. Here in this question according to given information option 4 may be

    correct answer.

$1. {\mathrm{C}}_{\mathrm{V}}=\frac{\mathrm{f}}{2}\mathrm{R}$  For monoatomic gas $\mathrm{f}=3 \Rightarrow {\left({\mathrm{C}}_{\mathrm{V}}\right)}_{\mathrm{mono}}=\frac{3}{2}\mathrm{R}.$

1. ${\mathrm{C}}_{\mathrm{P}}-{\mathrm{C}}_{\mathrm{V}}=\mathrm{R}=2.\frac{\mathrm{cal}}{\mathrm{gm}-\mathrm{mol}-\mathrm{K}}$ which is correct for option 1 and 2. Further the ratio $\frac{{\mathrm{C}}_{\mathrm{P}}}{{\mathrm{C}}_{\mathrm{V}}}\left(=\mathrm{\gamma }\right)$ should

    equal to some standard value corresponding to that of either, mono, di, or triatomic gases. From this

    point of view option 1 is correct because ${\left(\frac{{\mathrm{C}}_{\mathrm{P}}}{{\mathrm{C}}_{\mathrm{V}}}\right)}_{\mathrm{mono}}=\frac{5}{3}$

1. ${\mathrm{C}}_{\mathrm{P}}>{\mathrm{C}}_{\mathrm{V}}$

$3. {\mathrm{C}}_{\mathrm{isothermal}}=\infty \mathrm{and} {\mathrm{C}}_{\mathrm{adiabatic}}=0$

$$\begin{gathered} 3. \mathrm{\gamma }=1+\frac{2}{\mathrm{f}} \Rightarrow 1.4=1+\frac{2}{\mathrm{f}} \\ \Rightarrow \mathrm{Degree} \mathrm{of} \mathrm{freedom} \mathrm{f}=5 \\ \Rightarrow \mathrm{Degree} \mathrm{of} \mathrm{freedom} \mathrm{of} \mathrm{diatomic} \mathrm{gas} \mathrm{is} 5 \mathrm{and} \mathrm{it}\text{'}\mathrm{s} {\mathrm{C}}_{\mathrm{P}}=\frac{7}{2}\mathrm{R} \mathrm{and} {\mathrm{C}}_{\mathrm{V}}=\frac{5}{2}\mathrm{R} \end{gathered}$$

$$\begin{gathered} 4. {\mathrm{C}}_{\mathrm{P}}-{\mathrm{C}}_{\mathrm{V}}=\mathrm{R}=4150 \frac{\mathrm{J}}{\mathrm{kg}-\mathrm{K}} \mathrm{and} \frac{{\mathrm{C}}_{\mathrm{P}}}{{\mathrm{C}}_{\mathrm{V}}}=\mathrm{\gamma }=1.4 \\ \Rightarrow {\mathrm{C}}_{\mathrm{V}}=\frac{\mathrm{R}}{\left(\mathrm{\gamma }-1\right)}=\frac{4150}{\left(1.4-1\right)}=10375 \mathrm{J}/\mathrm{kg}-\mathrm{K} \end{gathered}$$

4. For any gas ${\mathrm{C}}_{\mathrm{P}}-{\mathrm{C}}_{\mathrm{V}}=1.99\approx 2\frac{\mathrm{cal}}{\mathrm{mol}-\mathrm{K}}$

1.

${\mathrm{\gamma }}_{\mathrm{mix}}=\frac{{\displaystyle \frac{{\mathrm{\mu }}_1{\mathrm{\gamma }}_1}{{\mathrm{\gamma }}_1-1}}+{\displaystyle \frac{{\mathrm{\mu }}_2{\mathrm{\gamma }}_2}{{\mathrm{\gamma }}_2-1}}}{{\displaystyle \frac{{\mathrm{\mu }}_1}{{\mathrm{\gamma }}_1-1}}+{\displaystyle \frac{{\mathrm{\mu }}_2}{{\mathrm{\gamma }}_2-1}}}=\frac{{\displaystyle \frac{1\times {\displaystyle \frac{5}{3}}}{\left({\displaystyle \frac{5}{3}}-1\right)}}+{\displaystyle \frac{1\times {\displaystyle \frac{7}{5}}}{\left({\displaystyle \frac{7}{5}}-1\right)}}}{{\displaystyle \frac{1}{\left({\displaystyle \frac{5}{3}}-1\right)}}+{\displaystyle \frac{1}{\left({\displaystyle \frac{7}{5}}-1\right)}}}=\frac{3}{2}=1.5$

3. Below 100 K only translational degree of freedom is considered. Hence 

$$\begin{gathered} {\mathrm{\gamma }}_{\mathrm{mixture}} = \frac{{\mathrm{\mu }}_1{\mathrm{C}}_{\mathrm{p},1 }+ {\mathrm{\mu }}_2{\mathrm{C}}_{\mathrm{p},2 }}{{\mathrm{\mu }}_1{\mathrm{C}}_{\mathrm{v},1 }+ {\mathrm{\mu }}_2{\mathrm{C}}_{\mathrm{v},2 }} \\ \mathrm{where} {\mathrm{\mu }}_1 \mathrm{and} {\mathrm{\mu }}_2 \mathrm{are} \mathrm{moles} \mathrm{of} \mathrm{samples} \mathrm{A} \mathrm{and} \mathrm{B} \\ {\mathrm{\gamma }}_{\mathrm{mixture}}=\frac{{\displaystyle \frac{{\mathrm{\mu }}_1{\mathrm{\gamma }}_1}{{\mathrm{\gamma }}_1-1}}+{\displaystyle \frac{{\mathrm{\mu }}_2{\mathrm{\gamma }}_2}{{\mathrm{\gamma }}_2-1}}}{{\displaystyle \frac{{\mathrm{\mu }}_1}{{\mathrm{\gamma }}_1-1}}+{\displaystyle \frac{{\mathrm{\mu }}_2}{{\mathrm{\gamma }}_2-1}}} \end{gathered}$$ 

according to question, 

${\mathrm{\mu }}_1={\mathrm{\mu }}_2 \mathrm{and} {\mathrm{\gamma }}_1={\mathrm{\gamma }}_2 =1+\frac{2}{3}=\frac{5}{3}$

$\Rightarrow {\mathrm{\gamma }}_{\mathrm{mix}}={\mathrm{\gamma }}_1=\frac{5}{3}$