NEET Test Series from KOTA - 10 Papers In MS WORD
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Kinetic Theory of Gases
139326
One mole of a monoatomic gas $\left(\gamma=\frac{5}{3}\right)$ is mixed with one mole of a diatomic gas $\left(\gamma=\frac{7}{5}\right)$. The value of $\gamma$ for the mixture is
1 1.40
2 1.50
3 1.53
4 3.0
Explanation:
B Given that $\mathrm{n}_{1}=1, \gamma_{1}=\frac{5}{3}$ $\mathrm{n}_{2}=1, \gamma_{2}=\frac{7}{5}$ $\gamma_{\text {mix }}=?$ We know that, $\frac{\mathrm{n}_{1}+\mathrm{n}_{2}}{\gamma_{\text {mix }}-1}=\frac{\mathrm{n}_{1}}{\gamma_{1}-1}+\frac{\mathrm{n}_{2}}{\gamma_{2}-1}=\frac{1}{\frac{5}{3}-1}+\frac{1}{\frac{7}{5}-1}$ $\frac{2}{\gamma_{\text {mix }}-1}=\frac{1}{\frac{5}{3}-1}+\frac{1}{\frac{7}{5}-1}$ $\frac{2}{\gamma_{\text {mix }}-1}=\frac{3}{2}+\frac{5}{2}$ $\frac{1}{\gamma_{\text {mix }}-1}=\frac{4}{2}$ $4 \gamma_{\text {mix }}-4=2$ $4 \gamma_{\text {mix }}=6$ $\gamma_{\text {mix }}=\frac{6}{4}=\frac{3}{2}$ $\gamma_{\text {mix }}=1.5$
UPSEE 2019
Kinetic Theory of Gases
139328
The value of $\gamma\left(=\frac{C_{p}}{C_{v}}\right)$, for hydrogen, helium and another ideal diatomic gas $\mathrm{X}$ (whose molecules are not rigid but have an additional vibrational mode), are respectively equal to
1 $\frac{7}{5}, \frac{5}{3}, \frac{9}{7}$
2 $\frac{5}{3}, \frac{7}{5}, \frac{9}{7}$
3 $\frac{5}{3}, \frac{7}{5}, \frac{7}{5}$
4 $\frac{7}{5}, \frac{5}{3}, \frac{7}{5}$
Explanation:
A We know that, Degree of freedom (n)- For mono atomic $(\mathrm{n})=1+\frac{2}{\mathrm{n}}=1+\frac{2}{3}$ We know that, $\mathrm{H}_{2}=\text { diatomic gas }$ $\mathrm{He}=\text { monoatomic gas }$ Diatomic gases have 5 degrees of freedom, if vibrational mode is neglected. If vibrational mode is also considered then degree of freedom of diatomic gas molecules are 7. Degrees of freedom of monoatomic gas is 3 . Hence, For Hydrogen, $(\gamma)=1+\frac{2}{\mathrm{f}} \quad(\mathrm{f}=$ degree of freedom $)$ $\gamma_{\mathrm{H}_{2}}=1+\frac{2}{5}=\frac{7}{5}$ For Helium, $(\gamma)=1+\frac{2}{f} \quad(\mathrm{f}=3)$ $\gamma_{\mathrm{He}}=1+\frac{2}{3}=\frac{5}{3}$ For gas $\mathrm{x}$ (vibration mode also considered) $\mathrm{f}=7$ $\gamma_{\mathrm{x}}=1+\frac{2}{\mathrm{f}}$ $\gamma_{\mathrm{x}}=1+\frac{2}{7}=\frac{9}{7}$ So, $\frac{7}{5}, \frac{5}{3}, \frac{9}{7}$
NEET Odisha-2019
Kinetic Theory of Gases
139329
Assertion: The ratio $\frac{C_{v}}{C_{p}}$ for a monatomic gas is less than one for a diatomic gas. Reason: The molecules of a monatomic gas have more degrees of freedom that those of a diatomic gas.
1 If both assertion and reason are true and reason is the correct explanation of assertion.
2 If both assertion and reason are true but reason is not the correct explanation of assertion.
3 If assertion is true but reason is false.
4 If both assertion and reason are false.
Explanation:
C For diatomic gas, degree of freedom $\mathrm{n}=5$ and $\mathrm{C}_{\mathrm{V}_{1}}=\frac{5}{2} \mathrm{R}, \quad \mathrm{C}_{\mathrm{P}_{1}}=\frac{7}{2} \mathrm{R}$ $\frac{\mathrm{C}_{\mathrm{V}_{1}}}{\mathrm{C}_{\mathrm{P}_{1}}}=\frac{5}{7}=0.714$ For monoatomic gas, degree of freedom $\mathrm{n}=3$ Hence, if assertion is true but reason is false.
AIIMS-26.05.2019(M) Shift-1
Kinetic Theory of Gases
139330
If $7 \mathrm{~g} \mathrm{~N}_{2}$ is mixed with $20 \mathrm{~g} \mathrm{Ar}$, then $\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{v}}$ of mixture will be:
1 $\frac{17}{6}$
2 $\frac{11}{7}$
3 $\frac{17}{11}$
4 $\frac{17}{13}$
Explanation:
C For $\mathrm{N}_{2}$, degree of freedom $\mathrm{f}_{1}=5$ Moles of $\mathrm{N}_{2}, \mathrm{n}_{1}=\frac{7}{28}=\frac{1}{4}$ $\mathrm{C}_{\mathrm{V}_{1}}=\frac{5}{2} \mathrm{R} \quad$ and $\mathrm{C}_{\mathrm{P}_{1}}=\frac{7}{2} \mathrm{R}$ For Ar, degree of freedom $\mathrm{f}_{2}=3$ No. of moles of $\mathrm{Ar}, \mathrm{n}_{2}=\frac{20}{40}=\frac{1}{2}$ For mixture $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma_{\text {mix }}=\frac{\mathrm{n}_{1} \mathrm{C}_{\mathrm{P}_{1}}+\mathrm{n}_{2} \mathrm{C}_{\mathrm{P}_{2}}}{\mathrm{n}_{1} \mathrm{C}_{\mathrm{V}_{1}}+\mathrm{n}_{2} \mathrm{C}_{\mathrm{V}_{2}}}$ Substituting the respective values $\gamma_{\text {mix }}=\frac{\frac{1}{4} \times \frac{7}{2} R+\frac{1}{2} \times \frac{5}{2} R}{\frac{1}{4} \times \frac{5}{2} R+\frac{1}{2} \times \frac{3}{2} R}=\frac{17}{11}$
139326
One mole of a monoatomic gas $\left(\gamma=\frac{5}{3}\right)$ is mixed with one mole of a diatomic gas $\left(\gamma=\frac{7}{5}\right)$. The value of $\gamma$ for the mixture is
1 1.40
2 1.50
3 1.53
4 3.0
Explanation:
B Given that $\mathrm{n}_{1}=1, \gamma_{1}=\frac{5}{3}$ $\mathrm{n}_{2}=1, \gamma_{2}=\frac{7}{5}$ $\gamma_{\text {mix }}=?$ We know that, $\frac{\mathrm{n}_{1}+\mathrm{n}_{2}}{\gamma_{\text {mix }}-1}=\frac{\mathrm{n}_{1}}{\gamma_{1}-1}+\frac{\mathrm{n}_{2}}{\gamma_{2}-1}=\frac{1}{\frac{5}{3}-1}+\frac{1}{\frac{7}{5}-1}$ $\frac{2}{\gamma_{\text {mix }}-1}=\frac{1}{\frac{5}{3}-1}+\frac{1}{\frac{7}{5}-1}$ $\frac{2}{\gamma_{\text {mix }}-1}=\frac{3}{2}+\frac{5}{2}$ $\frac{1}{\gamma_{\text {mix }}-1}=\frac{4}{2}$ $4 \gamma_{\text {mix }}-4=2$ $4 \gamma_{\text {mix }}=6$ $\gamma_{\text {mix }}=\frac{6}{4}=\frac{3}{2}$ $\gamma_{\text {mix }}=1.5$
UPSEE 2019
Kinetic Theory of Gases
139328
The value of $\gamma\left(=\frac{C_{p}}{C_{v}}\right)$, for hydrogen, helium and another ideal diatomic gas $\mathrm{X}$ (whose molecules are not rigid but have an additional vibrational mode), are respectively equal to
1 $\frac{7}{5}, \frac{5}{3}, \frac{9}{7}$
2 $\frac{5}{3}, \frac{7}{5}, \frac{9}{7}$
3 $\frac{5}{3}, \frac{7}{5}, \frac{7}{5}$
4 $\frac{7}{5}, \frac{5}{3}, \frac{7}{5}$
Explanation:
A We know that, Degree of freedom (n)- For mono atomic $(\mathrm{n})=1+\frac{2}{\mathrm{n}}=1+\frac{2}{3}$ We know that, $\mathrm{H}_{2}=\text { diatomic gas }$ $\mathrm{He}=\text { monoatomic gas }$ Diatomic gases have 5 degrees of freedom, if vibrational mode is neglected. If vibrational mode is also considered then degree of freedom of diatomic gas molecules are 7. Degrees of freedom of monoatomic gas is 3 . Hence, For Hydrogen, $(\gamma)=1+\frac{2}{\mathrm{f}} \quad(\mathrm{f}=$ degree of freedom $)$ $\gamma_{\mathrm{H}_{2}}=1+\frac{2}{5}=\frac{7}{5}$ For Helium, $(\gamma)=1+\frac{2}{f} \quad(\mathrm{f}=3)$ $\gamma_{\mathrm{He}}=1+\frac{2}{3}=\frac{5}{3}$ For gas $\mathrm{x}$ (vibration mode also considered) $\mathrm{f}=7$ $\gamma_{\mathrm{x}}=1+\frac{2}{\mathrm{f}}$ $\gamma_{\mathrm{x}}=1+\frac{2}{7}=\frac{9}{7}$ So, $\frac{7}{5}, \frac{5}{3}, \frac{9}{7}$
NEET Odisha-2019
Kinetic Theory of Gases
139329
Assertion: The ratio $\frac{C_{v}}{C_{p}}$ for a monatomic gas is less than one for a diatomic gas. Reason: The molecules of a monatomic gas have more degrees of freedom that those of a diatomic gas.
1 If both assertion and reason are true and reason is the correct explanation of assertion.
2 If both assertion and reason are true but reason is not the correct explanation of assertion.
3 If assertion is true but reason is false.
4 If both assertion and reason are false.
Explanation:
C For diatomic gas, degree of freedom $\mathrm{n}=5$ and $\mathrm{C}_{\mathrm{V}_{1}}=\frac{5}{2} \mathrm{R}, \quad \mathrm{C}_{\mathrm{P}_{1}}=\frac{7}{2} \mathrm{R}$ $\frac{\mathrm{C}_{\mathrm{V}_{1}}}{\mathrm{C}_{\mathrm{P}_{1}}}=\frac{5}{7}=0.714$ For monoatomic gas, degree of freedom $\mathrm{n}=3$ Hence, if assertion is true but reason is false.
AIIMS-26.05.2019(M) Shift-1
Kinetic Theory of Gases
139330
If $7 \mathrm{~g} \mathrm{~N}_{2}$ is mixed with $20 \mathrm{~g} \mathrm{Ar}$, then $\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{v}}$ of mixture will be:
1 $\frac{17}{6}$
2 $\frac{11}{7}$
3 $\frac{17}{11}$
4 $\frac{17}{13}$
Explanation:
C For $\mathrm{N}_{2}$, degree of freedom $\mathrm{f}_{1}=5$ Moles of $\mathrm{N}_{2}, \mathrm{n}_{1}=\frac{7}{28}=\frac{1}{4}$ $\mathrm{C}_{\mathrm{V}_{1}}=\frac{5}{2} \mathrm{R} \quad$ and $\mathrm{C}_{\mathrm{P}_{1}}=\frac{7}{2} \mathrm{R}$ For Ar, degree of freedom $\mathrm{f}_{2}=3$ No. of moles of $\mathrm{Ar}, \mathrm{n}_{2}=\frac{20}{40}=\frac{1}{2}$ For mixture $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma_{\text {mix }}=\frac{\mathrm{n}_{1} \mathrm{C}_{\mathrm{P}_{1}}+\mathrm{n}_{2} \mathrm{C}_{\mathrm{P}_{2}}}{\mathrm{n}_{1} \mathrm{C}_{\mathrm{V}_{1}}+\mathrm{n}_{2} \mathrm{C}_{\mathrm{V}_{2}}}$ Substituting the respective values $\gamma_{\text {mix }}=\frac{\frac{1}{4} \times \frac{7}{2} R+\frac{1}{2} \times \frac{5}{2} R}{\frac{1}{4} \times \frac{5}{2} R+\frac{1}{2} \times \frac{3}{2} R}=\frac{17}{11}$
139326
One mole of a monoatomic gas $\left(\gamma=\frac{5}{3}\right)$ is mixed with one mole of a diatomic gas $\left(\gamma=\frac{7}{5}\right)$. The value of $\gamma$ for the mixture is
1 1.40
2 1.50
3 1.53
4 3.0
Explanation:
B Given that $\mathrm{n}_{1}=1, \gamma_{1}=\frac{5}{3}$ $\mathrm{n}_{2}=1, \gamma_{2}=\frac{7}{5}$ $\gamma_{\text {mix }}=?$ We know that, $\frac{\mathrm{n}_{1}+\mathrm{n}_{2}}{\gamma_{\text {mix }}-1}=\frac{\mathrm{n}_{1}}{\gamma_{1}-1}+\frac{\mathrm{n}_{2}}{\gamma_{2}-1}=\frac{1}{\frac{5}{3}-1}+\frac{1}{\frac{7}{5}-1}$ $\frac{2}{\gamma_{\text {mix }}-1}=\frac{1}{\frac{5}{3}-1}+\frac{1}{\frac{7}{5}-1}$ $\frac{2}{\gamma_{\text {mix }}-1}=\frac{3}{2}+\frac{5}{2}$ $\frac{1}{\gamma_{\text {mix }}-1}=\frac{4}{2}$ $4 \gamma_{\text {mix }}-4=2$ $4 \gamma_{\text {mix }}=6$ $\gamma_{\text {mix }}=\frac{6}{4}=\frac{3}{2}$ $\gamma_{\text {mix }}=1.5$
UPSEE 2019
Kinetic Theory of Gases
139328
The value of $\gamma\left(=\frac{C_{p}}{C_{v}}\right)$, for hydrogen, helium and another ideal diatomic gas $\mathrm{X}$ (whose molecules are not rigid but have an additional vibrational mode), are respectively equal to
1 $\frac{7}{5}, \frac{5}{3}, \frac{9}{7}$
2 $\frac{5}{3}, \frac{7}{5}, \frac{9}{7}$
3 $\frac{5}{3}, \frac{7}{5}, \frac{7}{5}$
4 $\frac{7}{5}, \frac{5}{3}, \frac{7}{5}$
Explanation:
A We know that, Degree of freedom (n)- For mono atomic $(\mathrm{n})=1+\frac{2}{\mathrm{n}}=1+\frac{2}{3}$ We know that, $\mathrm{H}_{2}=\text { diatomic gas }$ $\mathrm{He}=\text { monoatomic gas }$ Diatomic gases have 5 degrees of freedom, if vibrational mode is neglected. If vibrational mode is also considered then degree of freedom of diatomic gas molecules are 7. Degrees of freedom of monoatomic gas is 3 . Hence, For Hydrogen, $(\gamma)=1+\frac{2}{\mathrm{f}} \quad(\mathrm{f}=$ degree of freedom $)$ $\gamma_{\mathrm{H}_{2}}=1+\frac{2}{5}=\frac{7}{5}$ For Helium, $(\gamma)=1+\frac{2}{f} \quad(\mathrm{f}=3)$ $\gamma_{\mathrm{He}}=1+\frac{2}{3}=\frac{5}{3}$ For gas $\mathrm{x}$ (vibration mode also considered) $\mathrm{f}=7$ $\gamma_{\mathrm{x}}=1+\frac{2}{\mathrm{f}}$ $\gamma_{\mathrm{x}}=1+\frac{2}{7}=\frac{9}{7}$ So, $\frac{7}{5}, \frac{5}{3}, \frac{9}{7}$
NEET Odisha-2019
Kinetic Theory of Gases
139329
Assertion: The ratio $\frac{C_{v}}{C_{p}}$ for a monatomic gas is less than one for a diatomic gas. Reason: The molecules of a monatomic gas have more degrees of freedom that those of a diatomic gas.
1 If both assertion and reason are true and reason is the correct explanation of assertion.
2 If both assertion and reason are true but reason is not the correct explanation of assertion.
3 If assertion is true but reason is false.
4 If both assertion and reason are false.
Explanation:
C For diatomic gas, degree of freedom $\mathrm{n}=5$ and $\mathrm{C}_{\mathrm{V}_{1}}=\frac{5}{2} \mathrm{R}, \quad \mathrm{C}_{\mathrm{P}_{1}}=\frac{7}{2} \mathrm{R}$ $\frac{\mathrm{C}_{\mathrm{V}_{1}}}{\mathrm{C}_{\mathrm{P}_{1}}}=\frac{5}{7}=0.714$ For monoatomic gas, degree of freedom $\mathrm{n}=3$ Hence, if assertion is true but reason is false.
AIIMS-26.05.2019(M) Shift-1
Kinetic Theory of Gases
139330
If $7 \mathrm{~g} \mathrm{~N}_{2}$ is mixed with $20 \mathrm{~g} \mathrm{Ar}$, then $\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{v}}$ of mixture will be:
1 $\frac{17}{6}$
2 $\frac{11}{7}$
3 $\frac{17}{11}$
4 $\frac{17}{13}$
Explanation:
C For $\mathrm{N}_{2}$, degree of freedom $\mathrm{f}_{1}=5$ Moles of $\mathrm{N}_{2}, \mathrm{n}_{1}=\frac{7}{28}=\frac{1}{4}$ $\mathrm{C}_{\mathrm{V}_{1}}=\frac{5}{2} \mathrm{R} \quad$ and $\mathrm{C}_{\mathrm{P}_{1}}=\frac{7}{2} \mathrm{R}$ For Ar, degree of freedom $\mathrm{f}_{2}=3$ No. of moles of $\mathrm{Ar}, \mathrm{n}_{2}=\frac{20}{40}=\frac{1}{2}$ For mixture $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma_{\text {mix }}=\frac{\mathrm{n}_{1} \mathrm{C}_{\mathrm{P}_{1}}+\mathrm{n}_{2} \mathrm{C}_{\mathrm{P}_{2}}}{\mathrm{n}_{1} \mathrm{C}_{\mathrm{V}_{1}}+\mathrm{n}_{2} \mathrm{C}_{\mathrm{V}_{2}}}$ Substituting the respective values $\gamma_{\text {mix }}=\frac{\frac{1}{4} \times \frac{7}{2} R+\frac{1}{2} \times \frac{5}{2} R}{\frac{1}{4} \times \frac{5}{2} R+\frac{1}{2} \times \frac{3}{2} R}=\frac{17}{11}$
139326
One mole of a monoatomic gas $\left(\gamma=\frac{5}{3}\right)$ is mixed with one mole of a diatomic gas $\left(\gamma=\frac{7}{5}\right)$. The value of $\gamma$ for the mixture is
1 1.40
2 1.50
3 1.53
4 3.0
Explanation:
B Given that $\mathrm{n}_{1}=1, \gamma_{1}=\frac{5}{3}$ $\mathrm{n}_{2}=1, \gamma_{2}=\frac{7}{5}$ $\gamma_{\text {mix }}=?$ We know that, $\frac{\mathrm{n}_{1}+\mathrm{n}_{2}}{\gamma_{\text {mix }}-1}=\frac{\mathrm{n}_{1}}{\gamma_{1}-1}+\frac{\mathrm{n}_{2}}{\gamma_{2}-1}=\frac{1}{\frac{5}{3}-1}+\frac{1}{\frac{7}{5}-1}$ $\frac{2}{\gamma_{\text {mix }}-1}=\frac{1}{\frac{5}{3}-1}+\frac{1}{\frac{7}{5}-1}$ $\frac{2}{\gamma_{\text {mix }}-1}=\frac{3}{2}+\frac{5}{2}$ $\frac{1}{\gamma_{\text {mix }}-1}=\frac{4}{2}$ $4 \gamma_{\text {mix }}-4=2$ $4 \gamma_{\text {mix }}=6$ $\gamma_{\text {mix }}=\frac{6}{4}=\frac{3}{2}$ $\gamma_{\text {mix }}=1.5$
UPSEE 2019
Kinetic Theory of Gases
139328
The value of $\gamma\left(=\frac{C_{p}}{C_{v}}\right)$, for hydrogen, helium and another ideal diatomic gas $\mathrm{X}$ (whose molecules are not rigid but have an additional vibrational mode), are respectively equal to
1 $\frac{7}{5}, \frac{5}{3}, \frac{9}{7}$
2 $\frac{5}{3}, \frac{7}{5}, \frac{9}{7}$
3 $\frac{5}{3}, \frac{7}{5}, \frac{7}{5}$
4 $\frac{7}{5}, \frac{5}{3}, \frac{7}{5}$
Explanation:
A We know that, Degree of freedom (n)- For mono atomic $(\mathrm{n})=1+\frac{2}{\mathrm{n}}=1+\frac{2}{3}$ We know that, $\mathrm{H}_{2}=\text { diatomic gas }$ $\mathrm{He}=\text { monoatomic gas }$ Diatomic gases have 5 degrees of freedom, if vibrational mode is neglected. If vibrational mode is also considered then degree of freedom of diatomic gas molecules are 7. Degrees of freedom of monoatomic gas is 3 . Hence, For Hydrogen, $(\gamma)=1+\frac{2}{\mathrm{f}} \quad(\mathrm{f}=$ degree of freedom $)$ $\gamma_{\mathrm{H}_{2}}=1+\frac{2}{5}=\frac{7}{5}$ For Helium, $(\gamma)=1+\frac{2}{f} \quad(\mathrm{f}=3)$ $\gamma_{\mathrm{He}}=1+\frac{2}{3}=\frac{5}{3}$ For gas $\mathrm{x}$ (vibration mode also considered) $\mathrm{f}=7$ $\gamma_{\mathrm{x}}=1+\frac{2}{\mathrm{f}}$ $\gamma_{\mathrm{x}}=1+\frac{2}{7}=\frac{9}{7}$ So, $\frac{7}{5}, \frac{5}{3}, \frac{9}{7}$
NEET Odisha-2019
Kinetic Theory of Gases
139329
Assertion: The ratio $\frac{C_{v}}{C_{p}}$ for a monatomic gas is less than one for a diatomic gas. Reason: The molecules of a monatomic gas have more degrees of freedom that those of a diatomic gas.
1 If both assertion and reason are true and reason is the correct explanation of assertion.
2 If both assertion and reason are true but reason is not the correct explanation of assertion.
3 If assertion is true but reason is false.
4 If both assertion and reason are false.
Explanation:
C For diatomic gas, degree of freedom $\mathrm{n}=5$ and $\mathrm{C}_{\mathrm{V}_{1}}=\frac{5}{2} \mathrm{R}, \quad \mathrm{C}_{\mathrm{P}_{1}}=\frac{7}{2} \mathrm{R}$ $\frac{\mathrm{C}_{\mathrm{V}_{1}}}{\mathrm{C}_{\mathrm{P}_{1}}}=\frac{5}{7}=0.714$ For monoatomic gas, degree of freedom $\mathrm{n}=3$ Hence, if assertion is true but reason is false.
AIIMS-26.05.2019(M) Shift-1
Kinetic Theory of Gases
139330
If $7 \mathrm{~g} \mathrm{~N}_{2}$ is mixed with $20 \mathrm{~g} \mathrm{Ar}$, then $\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{v}}$ of mixture will be:
1 $\frac{17}{6}$
2 $\frac{11}{7}$
3 $\frac{17}{11}$
4 $\frac{17}{13}$
Explanation:
C For $\mathrm{N}_{2}$, degree of freedom $\mathrm{f}_{1}=5$ Moles of $\mathrm{N}_{2}, \mathrm{n}_{1}=\frac{7}{28}=\frac{1}{4}$ $\mathrm{C}_{\mathrm{V}_{1}}=\frac{5}{2} \mathrm{R} \quad$ and $\mathrm{C}_{\mathrm{P}_{1}}=\frac{7}{2} \mathrm{R}$ For Ar, degree of freedom $\mathrm{f}_{2}=3$ No. of moles of $\mathrm{Ar}, \mathrm{n}_{2}=\frac{20}{40}=\frac{1}{2}$ For mixture $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma_{\text {mix }}=\frac{\mathrm{n}_{1} \mathrm{C}_{\mathrm{P}_{1}}+\mathrm{n}_{2} \mathrm{C}_{\mathrm{P}_{2}}}{\mathrm{n}_{1} \mathrm{C}_{\mathrm{V}_{1}}+\mathrm{n}_{2} \mathrm{C}_{\mathrm{V}_{2}}}$ Substituting the respective values $\gamma_{\text {mix }}=\frac{\frac{1}{4} \times \frac{7}{2} R+\frac{1}{2} \times \frac{5}{2} R}{\frac{1}{4} \times \frac{5}{2} R+\frac{1}{2} \times \frac{3}{2} R}=\frac{17}{11}$
