139313
If one mole of a monoatomic gas $\left(\gamma=\frac{5}{3}\right)$ is mixed with one mole 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 3.07
Explanation:
B Given that, For monoatomic gas $\gamma=\frac{5}{3}$ For diatomic gas $\gamma=\frac{7}{5}$ Now, for a monoatomic gas - $\mathrm{C}_{\mathrm{P}}=\frac{5 \mathrm{R}}{2} \text { and } \mathrm{C}_{\mathrm{V}}=\frac{3 \mathrm{R}}{2}$ For a diatomic gas $\mathrm{C}_{\mathrm{P}}=\frac{7 \mathrm{R}}{2} \text { and } \mathrm{C}_{\mathrm{V}}=\frac{5 \mathrm{R}}{2}$ So, for a mixture of 1 mole of each gas in solution - $\mathrm{C}_{\mathrm{V}_{\text {mix }}}=\frac{\frac{3 \mathrm{R}}{2}+\frac{5 \mathrm{R}}{2}}{2}=2 \mathrm{R}$ $\mathrm{C}_{\mathrm{P}_{\text {mix }}}=\frac{\frac{5 \mathrm{R}}{2}+\frac{7 \mathrm{R}}{2}}{2}=3 \mathrm{R}$ $\gamma_{\text {mix }}=\frac{\mathrm{C}_{\mathrm{P}_{\text {mix }}}}{\mathrm{C}_{\mathrm{V}_{\text {mix }}}}$ $\gamma_{\text {mix }}=\frac{3 \mathrm{R}}{2 \mathrm{R}}=1.5$
CG PET- 2015
Kinetic Theory of Gases
139314
2 moles of a diatomic gas are mixed with 1 mole of a monatomic gas. The ratio of two specific heats $\left(\gamma=C_{p} / C_{v}\right)$ of the mixture will be
1 $\frac{7}{3}$
2 $\frac{5}{4}$
3 $\frac{19}{13}$
4 $\frac{15}{19}$
Explanation:
C We know that, For diatomic gas $\mathrm{C}_{\mathrm{V}_{1}}=\frac{5}{2} \mathrm{R}$ and $\mathrm{C}_{\mathrm{P}_{1}}=\frac{7}{2} \mathrm{R}$ So, $\quad \gamma_{1}=\frac{\mathrm{C}_{\mathrm{P}_{1}}}{\mathrm{C}_{\mathrm{V}_{1}}}=\frac{\frac{7}{2} \mathrm{R}}{\frac{5}{2} \mathrm{R}}=\frac{7}{5}$ $\gamma_{1}=\frac{7}{5}$ Now, For monoatomic gas $\mathrm{C}_{\mathrm{V}_{2}}=\frac{3 \mathrm{R}}{2}$ and $\mathrm{C}_{\mathrm{P}_{2}}=\frac{5 \mathrm{R}}{2}$ $\gamma_{2}=\frac{\mathrm{C}_{\mathrm{P}_{2}}}{\mathrm{C}_{\mathrm{V}_{2}}}=\frac{\frac{5}{2} \mathrm{R}}{\frac{3 \mathrm{R}}{2}}$ $\gamma_{2}=\frac{5}{3}$ Let, $\gamma_{\text {mix }}$ is ratio of two specific heats of the mixture. $\gamma_{\text {mix }}=\frac{\frac{\mathrm{n}_{1} \gamma_{1}}{\gamma_{1}-1}+\frac{\mathrm{n}_{2} \gamma_{2}}{\gamma_{2}-1}}{\frac{\mathrm{n}_{1}}{\gamma_{1}-1}+\frac{\mathrm{n}_{2}}{\gamma_{2}-1}}$ Here, $\mathrm{n}_{1}=2$ mole, $\mathrm{n}_{2}=1$ mole $\frac{\frac{2 \times \frac{7}{5}}{\left(\frac{7}{5}-1\right)}+\frac{1 \times \frac{5}{3}}{\left(\frac{5}{3}-1\right)}}{\left(\frac{7}{5}-1\right)}+\frac{1}{\left(\frac{5}{3}-1\right)}$ $\gamma_{\text {mix }}= \frac{7+\frac{5}{2}}{\frac{10}{2}+\frac{3}{2}}=\frac{\frac{19}{2}}{\frac{7}{5}}$ $= \frac{19}{13}$
CG PET- 2011
Kinetic Theory of Gases
139317
For a certain gas the ratio of specific heats is given to be $r=1.5$. For this gas
A Given that, $\gamma=1.5=\frac{3}{2}$ We know that $\frac{C_{p}}{C_{v}}=\gamma$ $\frac{C_{p}}{C_{v}}=\frac{3}{2}$ $C_{V}=\frac{2}{3} C_{p}$ Again, $C_{p}-C_{v}=\frac{R}{J}$ $C_{p}-\frac{2}{3} C_{p}=\frac{R}{J}$ $\frac{C_{p}}{3}=\frac{R}{J}$ $C_{p}=\frac{3 R}{J}$
CG PET-2021
Kinetic Theory of Gases
139321
Two metallic spheres having same heat capacities are of radii in the ratio $2: 3$ and densities in the ratio $5: 6$. The ratio of their specific heats is nearly
1 $4: 1$
2 $2: 3$
3 $5: 4$
4 $=\frac{7}{5}$
Explanation:
A Given that, Ratio of radii of sphere, $\frac{r_{1}}{r_{2}}=\frac{2}{3}$ Ratio of density, $\frac{\rho_{1}}{\rho_{2}}=\frac{5}{6}$ For, Same heat capacities - $\mathrm{C}_{1}=\mathrm{C}_{2}$ $\mathrm{~m}_{1} \mathrm{c}_{1}=\mathrm{m}_{2} \mathrm{c}_{2}$ $\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}=\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}}$ $\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}=\frac{\mathrm{V}_{2} \cdot \rho_{2}}{\mathrm{~V}_{1} \cdot \rho_{1}}$ $\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}=\frac{\frac{4}{3} \pi \mathrm{r}_{2}^{3}}{\frac{4}{3} \pi \mathrm{r}_{1}^{3}} \times \frac{\rho_{2}}{\rho_{1}}$ $\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}=\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)^{3} \times \frac{\rho_{2}}{\rho_{1}}=\left(\frac{3}{2}\right)^{3} \times \frac{6}{5}$ $\frac{\mathrm{c}_{2}}{\mathrm{c}_{1}}=\frac{27}{8} \times \frac{6}{5}=\frac{81}{20}$ $\frac{\mathrm{c}_{2}}{\mathrm{c}_{1}}=\frac{4}{1}$
AP EAMCET (Medical)-07.10.2020
Kinetic Theory of Gases
139316
In non-rigid diatomic molecule with an additional vibrational mode #[Qdiff: Hard, QCat: Numerical Based, examname: AP EAMCET-04.07.2022,Shift-II#
139313
If one mole of a monoatomic gas $\left(\gamma=\frac{5}{3}\right)$ is mixed with one mole 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 3.07
Explanation:
B Given that, For monoatomic gas $\gamma=\frac{5}{3}$ For diatomic gas $\gamma=\frac{7}{5}$ Now, for a monoatomic gas - $\mathrm{C}_{\mathrm{P}}=\frac{5 \mathrm{R}}{2} \text { and } \mathrm{C}_{\mathrm{V}}=\frac{3 \mathrm{R}}{2}$ For a diatomic gas $\mathrm{C}_{\mathrm{P}}=\frac{7 \mathrm{R}}{2} \text { and } \mathrm{C}_{\mathrm{V}}=\frac{5 \mathrm{R}}{2}$ So, for a mixture of 1 mole of each gas in solution - $\mathrm{C}_{\mathrm{V}_{\text {mix }}}=\frac{\frac{3 \mathrm{R}}{2}+\frac{5 \mathrm{R}}{2}}{2}=2 \mathrm{R}$ $\mathrm{C}_{\mathrm{P}_{\text {mix }}}=\frac{\frac{5 \mathrm{R}}{2}+\frac{7 \mathrm{R}}{2}}{2}=3 \mathrm{R}$ $\gamma_{\text {mix }}=\frac{\mathrm{C}_{\mathrm{P}_{\text {mix }}}}{\mathrm{C}_{\mathrm{V}_{\text {mix }}}}$ $\gamma_{\text {mix }}=\frac{3 \mathrm{R}}{2 \mathrm{R}}=1.5$
CG PET- 2015
Kinetic Theory of Gases
139314
2 moles of a diatomic gas are mixed with 1 mole of a monatomic gas. The ratio of two specific heats $\left(\gamma=C_{p} / C_{v}\right)$ of the mixture will be
1 $\frac{7}{3}$
2 $\frac{5}{4}$
3 $\frac{19}{13}$
4 $\frac{15}{19}$
Explanation:
C We know that, For diatomic gas $\mathrm{C}_{\mathrm{V}_{1}}=\frac{5}{2} \mathrm{R}$ and $\mathrm{C}_{\mathrm{P}_{1}}=\frac{7}{2} \mathrm{R}$ So, $\quad \gamma_{1}=\frac{\mathrm{C}_{\mathrm{P}_{1}}}{\mathrm{C}_{\mathrm{V}_{1}}}=\frac{\frac{7}{2} \mathrm{R}}{\frac{5}{2} \mathrm{R}}=\frac{7}{5}$ $\gamma_{1}=\frac{7}{5}$ Now, For monoatomic gas $\mathrm{C}_{\mathrm{V}_{2}}=\frac{3 \mathrm{R}}{2}$ and $\mathrm{C}_{\mathrm{P}_{2}}=\frac{5 \mathrm{R}}{2}$ $\gamma_{2}=\frac{\mathrm{C}_{\mathrm{P}_{2}}}{\mathrm{C}_{\mathrm{V}_{2}}}=\frac{\frac{5}{2} \mathrm{R}}{\frac{3 \mathrm{R}}{2}}$ $\gamma_{2}=\frac{5}{3}$ Let, $\gamma_{\text {mix }}$ is ratio of two specific heats of the mixture. $\gamma_{\text {mix }}=\frac{\frac{\mathrm{n}_{1} \gamma_{1}}{\gamma_{1}-1}+\frac{\mathrm{n}_{2} \gamma_{2}}{\gamma_{2}-1}}{\frac{\mathrm{n}_{1}}{\gamma_{1}-1}+\frac{\mathrm{n}_{2}}{\gamma_{2}-1}}$ Here, $\mathrm{n}_{1}=2$ mole, $\mathrm{n}_{2}=1$ mole $\frac{\frac{2 \times \frac{7}{5}}{\left(\frac{7}{5}-1\right)}+\frac{1 \times \frac{5}{3}}{\left(\frac{5}{3}-1\right)}}{\left(\frac{7}{5}-1\right)}+\frac{1}{\left(\frac{5}{3}-1\right)}$ $\gamma_{\text {mix }}= \frac{7+\frac{5}{2}}{\frac{10}{2}+\frac{3}{2}}=\frac{\frac{19}{2}}{\frac{7}{5}}$ $= \frac{19}{13}$
CG PET- 2011
Kinetic Theory of Gases
139317
For a certain gas the ratio of specific heats is given to be $r=1.5$. For this gas
A Given that, $\gamma=1.5=\frac{3}{2}$ We know that $\frac{C_{p}}{C_{v}}=\gamma$ $\frac{C_{p}}{C_{v}}=\frac{3}{2}$ $C_{V}=\frac{2}{3} C_{p}$ Again, $C_{p}-C_{v}=\frac{R}{J}$ $C_{p}-\frac{2}{3} C_{p}=\frac{R}{J}$ $\frac{C_{p}}{3}=\frac{R}{J}$ $C_{p}=\frac{3 R}{J}$
CG PET-2021
Kinetic Theory of Gases
139321
Two metallic spheres having same heat capacities are of radii in the ratio $2: 3$ and densities in the ratio $5: 6$. The ratio of their specific heats is nearly
1 $4: 1$
2 $2: 3$
3 $5: 4$
4 $=\frac{7}{5}$
Explanation:
A Given that, Ratio of radii of sphere, $\frac{r_{1}}{r_{2}}=\frac{2}{3}$ Ratio of density, $\frac{\rho_{1}}{\rho_{2}}=\frac{5}{6}$ For, Same heat capacities - $\mathrm{C}_{1}=\mathrm{C}_{2}$ $\mathrm{~m}_{1} \mathrm{c}_{1}=\mathrm{m}_{2} \mathrm{c}_{2}$ $\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}=\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}}$ $\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}=\frac{\mathrm{V}_{2} \cdot \rho_{2}}{\mathrm{~V}_{1} \cdot \rho_{1}}$ $\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}=\frac{\frac{4}{3} \pi \mathrm{r}_{2}^{3}}{\frac{4}{3} \pi \mathrm{r}_{1}^{3}} \times \frac{\rho_{2}}{\rho_{1}}$ $\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}=\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)^{3} \times \frac{\rho_{2}}{\rho_{1}}=\left(\frac{3}{2}\right)^{3} \times \frac{6}{5}$ $\frac{\mathrm{c}_{2}}{\mathrm{c}_{1}}=\frac{27}{8} \times \frac{6}{5}=\frac{81}{20}$ $\frac{\mathrm{c}_{2}}{\mathrm{c}_{1}}=\frac{4}{1}$
AP EAMCET (Medical)-07.10.2020
Kinetic Theory of Gases
139316
In non-rigid diatomic molecule with an additional vibrational mode #[Qdiff: Hard, QCat: Numerical Based, examname: AP EAMCET-04.07.2022,Shift-II#
139313
If one mole of a monoatomic gas $\left(\gamma=\frac{5}{3}\right)$ is mixed with one mole 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 3.07
Explanation:
B Given that, For monoatomic gas $\gamma=\frac{5}{3}$ For diatomic gas $\gamma=\frac{7}{5}$ Now, for a monoatomic gas - $\mathrm{C}_{\mathrm{P}}=\frac{5 \mathrm{R}}{2} \text { and } \mathrm{C}_{\mathrm{V}}=\frac{3 \mathrm{R}}{2}$ For a diatomic gas $\mathrm{C}_{\mathrm{P}}=\frac{7 \mathrm{R}}{2} \text { and } \mathrm{C}_{\mathrm{V}}=\frac{5 \mathrm{R}}{2}$ So, for a mixture of 1 mole of each gas in solution - $\mathrm{C}_{\mathrm{V}_{\text {mix }}}=\frac{\frac{3 \mathrm{R}}{2}+\frac{5 \mathrm{R}}{2}}{2}=2 \mathrm{R}$ $\mathrm{C}_{\mathrm{P}_{\text {mix }}}=\frac{\frac{5 \mathrm{R}}{2}+\frac{7 \mathrm{R}}{2}}{2}=3 \mathrm{R}$ $\gamma_{\text {mix }}=\frac{\mathrm{C}_{\mathrm{P}_{\text {mix }}}}{\mathrm{C}_{\mathrm{V}_{\text {mix }}}}$ $\gamma_{\text {mix }}=\frac{3 \mathrm{R}}{2 \mathrm{R}}=1.5$
CG PET- 2015
Kinetic Theory of Gases
139314
2 moles of a diatomic gas are mixed with 1 mole of a monatomic gas. The ratio of two specific heats $\left(\gamma=C_{p} / C_{v}\right)$ of the mixture will be
1 $\frac{7}{3}$
2 $\frac{5}{4}$
3 $\frac{19}{13}$
4 $\frac{15}{19}$
Explanation:
C We know that, For diatomic gas $\mathrm{C}_{\mathrm{V}_{1}}=\frac{5}{2} \mathrm{R}$ and $\mathrm{C}_{\mathrm{P}_{1}}=\frac{7}{2} \mathrm{R}$ So, $\quad \gamma_{1}=\frac{\mathrm{C}_{\mathrm{P}_{1}}}{\mathrm{C}_{\mathrm{V}_{1}}}=\frac{\frac{7}{2} \mathrm{R}}{\frac{5}{2} \mathrm{R}}=\frac{7}{5}$ $\gamma_{1}=\frac{7}{5}$ Now, For monoatomic gas $\mathrm{C}_{\mathrm{V}_{2}}=\frac{3 \mathrm{R}}{2}$ and $\mathrm{C}_{\mathrm{P}_{2}}=\frac{5 \mathrm{R}}{2}$ $\gamma_{2}=\frac{\mathrm{C}_{\mathrm{P}_{2}}}{\mathrm{C}_{\mathrm{V}_{2}}}=\frac{\frac{5}{2} \mathrm{R}}{\frac{3 \mathrm{R}}{2}}$ $\gamma_{2}=\frac{5}{3}$ Let, $\gamma_{\text {mix }}$ is ratio of two specific heats of the mixture. $\gamma_{\text {mix }}=\frac{\frac{\mathrm{n}_{1} \gamma_{1}}{\gamma_{1}-1}+\frac{\mathrm{n}_{2} \gamma_{2}}{\gamma_{2}-1}}{\frac{\mathrm{n}_{1}}{\gamma_{1}-1}+\frac{\mathrm{n}_{2}}{\gamma_{2}-1}}$ Here, $\mathrm{n}_{1}=2$ mole, $\mathrm{n}_{2}=1$ mole $\frac{\frac{2 \times \frac{7}{5}}{\left(\frac{7}{5}-1\right)}+\frac{1 \times \frac{5}{3}}{\left(\frac{5}{3}-1\right)}}{\left(\frac{7}{5}-1\right)}+\frac{1}{\left(\frac{5}{3}-1\right)}$ $\gamma_{\text {mix }}= \frac{7+\frac{5}{2}}{\frac{10}{2}+\frac{3}{2}}=\frac{\frac{19}{2}}{\frac{7}{5}}$ $= \frac{19}{13}$
CG PET- 2011
Kinetic Theory of Gases
139317
For a certain gas the ratio of specific heats is given to be $r=1.5$. For this gas
A Given that, $\gamma=1.5=\frac{3}{2}$ We know that $\frac{C_{p}}{C_{v}}=\gamma$ $\frac{C_{p}}{C_{v}}=\frac{3}{2}$ $C_{V}=\frac{2}{3} C_{p}$ Again, $C_{p}-C_{v}=\frac{R}{J}$ $C_{p}-\frac{2}{3} C_{p}=\frac{R}{J}$ $\frac{C_{p}}{3}=\frac{R}{J}$ $C_{p}=\frac{3 R}{J}$
CG PET-2021
Kinetic Theory of Gases
139321
Two metallic spheres having same heat capacities are of radii in the ratio $2: 3$ and densities in the ratio $5: 6$. The ratio of their specific heats is nearly
1 $4: 1$
2 $2: 3$
3 $5: 4$
4 $=\frac{7}{5}$
Explanation:
A Given that, Ratio of radii of sphere, $\frac{r_{1}}{r_{2}}=\frac{2}{3}$ Ratio of density, $\frac{\rho_{1}}{\rho_{2}}=\frac{5}{6}$ For, Same heat capacities - $\mathrm{C}_{1}=\mathrm{C}_{2}$ $\mathrm{~m}_{1} \mathrm{c}_{1}=\mathrm{m}_{2} \mathrm{c}_{2}$ $\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}=\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}}$ $\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}=\frac{\mathrm{V}_{2} \cdot \rho_{2}}{\mathrm{~V}_{1} \cdot \rho_{1}}$ $\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}=\frac{\frac{4}{3} \pi \mathrm{r}_{2}^{3}}{\frac{4}{3} \pi \mathrm{r}_{1}^{3}} \times \frac{\rho_{2}}{\rho_{1}}$ $\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}=\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)^{3} \times \frac{\rho_{2}}{\rho_{1}}=\left(\frac{3}{2}\right)^{3} \times \frac{6}{5}$ $\frac{\mathrm{c}_{2}}{\mathrm{c}_{1}}=\frac{27}{8} \times \frac{6}{5}=\frac{81}{20}$ $\frac{\mathrm{c}_{2}}{\mathrm{c}_{1}}=\frac{4}{1}$
AP EAMCET (Medical)-07.10.2020
Kinetic Theory of Gases
139316
In non-rigid diatomic molecule with an additional vibrational mode #[Qdiff: Hard, QCat: Numerical Based, examname: AP EAMCET-04.07.2022,Shift-II#
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Kinetic Theory of Gases
139313
If one mole of a monoatomic gas $\left(\gamma=\frac{5}{3}\right)$ is mixed with one mole 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 3.07
Explanation:
B Given that, For monoatomic gas $\gamma=\frac{5}{3}$ For diatomic gas $\gamma=\frac{7}{5}$ Now, for a monoatomic gas - $\mathrm{C}_{\mathrm{P}}=\frac{5 \mathrm{R}}{2} \text { and } \mathrm{C}_{\mathrm{V}}=\frac{3 \mathrm{R}}{2}$ For a diatomic gas $\mathrm{C}_{\mathrm{P}}=\frac{7 \mathrm{R}}{2} \text { and } \mathrm{C}_{\mathrm{V}}=\frac{5 \mathrm{R}}{2}$ So, for a mixture of 1 mole of each gas in solution - $\mathrm{C}_{\mathrm{V}_{\text {mix }}}=\frac{\frac{3 \mathrm{R}}{2}+\frac{5 \mathrm{R}}{2}}{2}=2 \mathrm{R}$ $\mathrm{C}_{\mathrm{P}_{\text {mix }}}=\frac{\frac{5 \mathrm{R}}{2}+\frac{7 \mathrm{R}}{2}}{2}=3 \mathrm{R}$ $\gamma_{\text {mix }}=\frac{\mathrm{C}_{\mathrm{P}_{\text {mix }}}}{\mathrm{C}_{\mathrm{V}_{\text {mix }}}}$ $\gamma_{\text {mix }}=\frac{3 \mathrm{R}}{2 \mathrm{R}}=1.5$
CG PET- 2015
Kinetic Theory of Gases
139314
2 moles of a diatomic gas are mixed with 1 mole of a monatomic gas. The ratio of two specific heats $\left(\gamma=C_{p} / C_{v}\right)$ of the mixture will be
1 $\frac{7}{3}$
2 $\frac{5}{4}$
3 $\frac{19}{13}$
4 $\frac{15}{19}$
Explanation:
C We know that, For diatomic gas $\mathrm{C}_{\mathrm{V}_{1}}=\frac{5}{2} \mathrm{R}$ and $\mathrm{C}_{\mathrm{P}_{1}}=\frac{7}{2} \mathrm{R}$ So, $\quad \gamma_{1}=\frac{\mathrm{C}_{\mathrm{P}_{1}}}{\mathrm{C}_{\mathrm{V}_{1}}}=\frac{\frac{7}{2} \mathrm{R}}{\frac{5}{2} \mathrm{R}}=\frac{7}{5}$ $\gamma_{1}=\frac{7}{5}$ Now, For monoatomic gas $\mathrm{C}_{\mathrm{V}_{2}}=\frac{3 \mathrm{R}}{2}$ and $\mathrm{C}_{\mathrm{P}_{2}}=\frac{5 \mathrm{R}}{2}$ $\gamma_{2}=\frac{\mathrm{C}_{\mathrm{P}_{2}}}{\mathrm{C}_{\mathrm{V}_{2}}}=\frac{\frac{5}{2} \mathrm{R}}{\frac{3 \mathrm{R}}{2}}$ $\gamma_{2}=\frac{5}{3}$ Let, $\gamma_{\text {mix }}$ is ratio of two specific heats of the mixture. $\gamma_{\text {mix }}=\frac{\frac{\mathrm{n}_{1} \gamma_{1}}{\gamma_{1}-1}+\frac{\mathrm{n}_{2} \gamma_{2}}{\gamma_{2}-1}}{\frac{\mathrm{n}_{1}}{\gamma_{1}-1}+\frac{\mathrm{n}_{2}}{\gamma_{2}-1}}$ Here, $\mathrm{n}_{1}=2$ mole, $\mathrm{n}_{2}=1$ mole $\frac{\frac{2 \times \frac{7}{5}}{\left(\frac{7}{5}-1\right)}+\frac{1 \times \frac{5}{3}}{\left(\frac{5}{3}-1\right)}}{\left(\frac{7}{5}-1\right)}+\frac{1}{\left(\frac{5}{3}-1\right)}$ $\gamma_{\text {mix }}= \frac{7+\frac{5}{2}}{\frac{10}{2}+\frac{3}{2}}=\frac{\frac{19}{2}}{\frac{7}{5}}$ $= \frac{19}{13}$
CG PET- 2011
Kinetic Theory of Gases
139317
For a certain gas the ratio of specific heats is given to be $r=1.5$. For this gas
A Given that, $\gamma=1.5=\frac{3}{2}$ We know that $\frac{C_{p}}{C_{v}}=\gamma$ $\frac{C_{p}}{C_{v}}=\frac{3}{2}$ $C_{V}=\frac{2}{3} C_{p}$ Again, $C_{p}-C_{v}=\frac{R}{J}$ $C_{p}-\frac{2}{3} C_{p}=\frac{R}{J}$ $\frac{C_{p}}{3}=\frac{R}{J}$ $C_{p}=\frac{3 R}{J}$
CG PET-2021
Kinetic Theory of Gases
139321
Two metallic spheres having same heat capacities are of radii in the ratio $2: 3$ and densities in the ratio $5: 6$. The ratio of their specific heats is nearly
1 $4: 1$
2 $2: 3$
3 $5: 4$
4 $=\frac{7}{5}$
Explanation:
A Given that, Ratio of radii of sphere, $\frac{r_{1}}{r_{2}}=\frac{2}{3}$ Ratio of density, $\frac{\rho_{1}}{\rho_{2}}=\frac{5}{6}$ For, Same heat capacities - $\mathrm{C}_{1}=\mathrm{C}_{2}$ $\mathrm{~m}_{1} \mathrm{c}_{1}=\mathrm{m}_{2} \mathrm{c}_{2}$ $\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}=\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}}$ $\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}=\frac{\mathrm{V}_{2} \cdot \rho_{2}}{\mathrm{~V}_{1} \cdot \rho_{1}}$ $\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}=\frac{\frac{4}{3} \pi \mathrm{r}_{2}^{3}}{\frac{4}{3} \pi \mathrm{r}_{1}^{3}} \times \frac{\rho_{2}}{\rho_{1}}$ $\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}=\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)^{3} \times \frac{\rho_{2}}{\rho_{1}}=\left(\frac{3}{2}\right)^{3} \times \frac{6}{5}$ $\frac{\mathrm{c}_{2}}{\mathrm{c}_{1}}=\frac{27}{8} \times \frac{6}{5}=\frac{81}{20}$ $\frac{\mathrm{c}_{2}}{\mathrm{c}_{1}}=\frac{4}{1}$
AP EAMCET (Medical)-07.10.2020
Kinetic Theory of Gases
139316
In non-rigid diatomic molecule with an additional vibrational mode #[Qdiff: Hard, QCat: Numerical Based, examname: AP EAMCET-04.07.2022,Shift-II#
139313
If one mole of a monoatomic gas $\left(\gamma=\frac{5}{3}\right)$ is mixed with one mole 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 3.07
Explanation:
B Given that, For monoatomic gas $\gamma=\frac{5}{3}$ For diatomic gas $\gamma=\frac{7}{5}$ Now, for a monoatomic gas - $\mathrm{C}_{\mathrm{P}}=\frac{5 \mathrm{R}}{2} \text { and } \mathrm{C}_{\mathrm{V}}=\frac{3 \mathrm{R}}{2}$ For a diatomic gas $\mathrm{C}_{\mathrm{P}}=\frac{7 \mathrm{R}}{2} \text { and } \mathrm{C}_{\mathrm{V}}=\frac{5 \mathrm{R}}{2}$ So, for a mixture of 1 mole of each gas in solution - $\mathrm{C}_{\mathrm{V}_{\text {mix }}}=\frac{\frac{3 \mathrm{R}}{2}+\frac{5 \mathrm{R}}{2}}{2}=2 \mathrm{R}$ $\mathrm{C}_{\mathrm{P}_{\text {mix }}}=\frac{\frac{5 \mathrm{R}}{2}+\frac{7 \mathrm{R}}{2}}{2}=3 \mathrm{R}$ $\gamma_{\text {mix }}=\frac{\mathrm{C}_{\mathrm{P}_{\text {mix }}}}{\mathrm{C}_{\mathrm{V}_{\text {mix }}}}$ $\gamma_{\text {mix }}=\frac{3 \mathrm{R}}{2 \mathrm{R}}=1.5$
CG PET- 2015
Kinetic Theory of Gases
139314
2 moles of a diatomic gas are mixed with 1 mole of a monatomic gas. The ratio of two specific heats $\left(\gamma=C_{p} / C_{v}\right)$ of the mixture will be
1 $\frac{7}{3}$
2 $\frac{5}{4}$
3 $\frac{19}{13}$
4 $\frac{15}{19}$
Explanation:
C We know that, For diatomic gas $\mathrm{C}_{\mathrm{V}_{1}}=\frac{5}{2} \mathrm{R}$ and $\mathrm{C}_{\mathrm{P}_{1}}=\frac{7}{2} \mathrm{R}$ So, $\quad \gamma_{1}=\frac{\mathrm{C}_{\mathrm{P}_{1}}}{\mathrm{C}_{\mathrm{V}_{1}}}=\frac{\frac{7}{2} \mathrm{R}}{\frac{5}{2} \mathrm{R}}=\frac{7}{5}$ $\gamma_{1}=\frac{7}{5}$ Now, For monoatomic gas $\mathrm{C}_{\mathrm{V}_{2}}=\frac{3 \mathrm{R}}{2}$ and $\mathrm{C}_{\mathrm{P}_{2}}=\frac{5 \mathrm{R}}{2}$ $\gamma_{2}=\frac{\mathrm{C}_{\mathrm{P}_{2}}}{\mathrm{C}_{\mathrm{V}_{2}}}=\frac{\frac{5}{2} \mathrm{R}}{\frac{3 \mathrm{R}}{2}}$ $\gamma_{2}=\frac{5}{3}$ Let, $\gamma_{\text {mix }}$ is ratio of two specific heats of the mixture. $\gamma_{\text {mix }}=\frac{\frac{\mathrm{n}_{1} \gamma_{1}}{\gamma_{1}-1}+\frac{\mathrm{n}_{2} \gamma_{2}}{\gamma_{2}-1}}{\frac{\mathrm{n}_{1}}{\gamma_{1}-1}+\frac{\mathrm{n}_{2}}{\gamma_{2}-1}}$ Here, $\mathrm{n}_{1}=2$ mole, $\mathrm{n}_{2}=1$ mole $\frac{\frac{2 \times \frac{7}{5}}{\left(\frac{7}{5}-1\right)}+\frac{1 \times \frac{5}{3}}{\left(\frac{5}{3}-1\right)}}{\left(\frac{7}{5}-1\right)}+\frac{1}{\left(\frac{5}{3}-1\right)}$ $\gamma_{\text {mix }}= \frac{7+\frac{5}{2}}{\frac{10}{2}+\frac{3}{2}}=\frac{\frac{19}{2}}{\frac{7}{5}}$ $= \frac{19}{13}$
CG PET- 2011
Kinetic Theory of Gases
139317
For a certain gas the ratio of specific heats is given to be $r=1.5$. For this gas
A Given that, $\gamma=1.5=\frac{3}{2}$ We know that $\frac{C_{p}}{C_{v}}=\gamma$ $\frac{C_{p}}{C_{v}}=\frac{3}{2}$ $C_{V}=\frac{2}{3} C_{p}$ Again, $C_{p}-C_{v}=\frac{R}{J}$ $C_{p}-\frac{2}{3} C_{p}=\frac{R}{J}$ $\frac{C_{p}}{3}=\frac{R}{J}$ $C_{p}=\frac{3 R}{J}$
CG PET-2021
Kinetic Theory of Gases
139321
Two metallic spheres having same heat capacities are of radii in the ratio $2: 3$ and densities in the ratio $5: 6$. The ratio of their specific heats is nearly
1 $4: 1$
2 $2: 3$
3 $5: 4$
4 $=\frac{7}{5}$
Explanation:
A Given that, Ratio of radii of sphere, $\frac{r_{1}}{r_{2}}=\frac{2}{3}$ Ratio of density, $\frac{\rho_{1}}{\rho_{2}}=\frac{5}{6}$ For, Same heat capacities - $\mathrm{C}_{1}=\mathrm{C}_{2}$ $\mathrm{~m}_{1} \mathrm{c}_{1}=\mathrm{m}_{2} \mathrm{c}_{2}$ $\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}=\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}}$ $\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}=\frac{\mathrm{V}_{2} \cdot \rho_{2}}{\mathrm{~V}_{1} \cdot \rho_{1}}$ $\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}=\frac{\frac{4}{3} \pi \mathrm{r}_{2}^{3}}{\frac{4}{3} \pi \mathrm{r}_{1}^{3}} \times \frac{\rho_{2}}{\rho_{1}}$ $\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}=\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)^{3} \times \frac{\rho_{2}}{\rho_{1}}=\left(\frac{3}{2}\right)^{3} \times \frac{6}{5}$ $\frac{\mathrm{c}_{2}}{\mathrm{c}_{1}}=\frac{27}{8} \times \frac{6}{5}=\frac{81}{20}$ $\frac{\mathrm{c}_{2}}{\mathrm{c}_{1}}=\frac{4}{1}$
AP EAMCET (Medical)-07.10.2020
Kinetic Theory of Gases
139316
In non-rigid diatomic molecule with an additional vibrational mode #[Qdiff: Hard, QCat: Numerical Based, examname: AP EAMCET-04.07.2022,Shift-II#