139122
A polyatomic gas has $f$ vibrational degrees of freedom, then the ratio of the specific heat at constant pressure to that at constant volume will be
1 $\frac{4+\mathrm{f}}{3+\mathrm{f}}$
2 $\frac{4-\mathrm{f}}{3-\mathrm{f}}$
3 $\frac{3+\mathrm{f}}{3+\mathrm{f}}$
4 $\frac{3-\mathrm{f}}{4-\mathrm{f}}$
Explanation:
A We know that, $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}$ $\mathrm{f}=$ vibrational degrees of freedom For polyatomic gas the specific heat at constant volume, $\mathrm{C}_{\mathrm{V}}=(3+\mathrm{f}) \mathrm{R}$ So, from equation (i), we get- $C_{P}=C_{V}+R$ $C_{P}=(3+f) R+R$ $C_{P}=(4+f) R$ Ratio of $\frac{C_{P}}{C_{V}}=\frac{(4+f) R}{(3+f) R}=\frac{(4+f)}{(3+f)}$
Shift-1]
Kinetic Theory of Gases
139123
The temperature at which the r.m.s. velocity of oxygen molecules will be same as that of hydrogen molecules at $-173^{\circ} \mathrm{C}$ is
1 $1600^{\circ} \mathrm{C}$
2 $373^{\circ} \mathrm{C}$
3 $1327^{\circ} \mathrm{C}$
4 $1200^{\circ} \mathrm{C}$
Explanation:
C We have, $\mathrm{M}_{\mathrm{O}_{2}}=32$ $\mathrm{M}_{\mathrm{H}_{2}}=2$ Temperature of hydrogen molecules $\mathrm{T}_{1}=273+\left(-173^{\circ}\right)$ $\mathrm{T}_{1}=100 \mathrm{~K}$ Temperature of oxygen molecules $=T_{2}$ $\left(\mathrm{V}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}=\sqrt{\frac{3 \mathrm{RT}_{2}}{\mathrm{M}_{\mathrm{O}_{2}}}}$ $\left(\mathrm{~V}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}=\sqrt{\frac{3 \mathrm{RT}_{1}}{\mathrm{M}_{\mathrm{H}_{2}}}}$ According to question $\left(\mathrm{V}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}=\left(\mathrm{V}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}$ $\sqrt{\frac{3 \mathrm{RT}_{2}}{\mathrm{M}_{\mathrm{O}_{2}}}}=\sqrt{\frac{3 \mathrm{RT}_{1}}{\mathrm{M}_{\mathrm{H}_{2}}}}$ $\frac{3 \times \mathrm{T}_{2} \times \mathrm{R}}{32}=\frac{3 \times 100 \times \mathrm{R}}{2}$ $3 \times \mathrm{T}_{2} \times \mathrm{R}=4800 \times \mathrm{R}$ $\mathrm{T}_{2}=1600 \mathrm{~K}$ $\mathrm{~T}_{2}=(1600-273)^{\circ} \mathrm{C}$ $\mathrm{T}_{2}=1327^{\circ} \mathrm{C}$
COMEDK 2019
Kinetic Theory of Gases
139124
If ' $f$ ' is the number of degrees of freedom of a molecule of a gas and ratio of molar specific heats of a gas, $\gamma=1+\frac{2}{f}$ where $\gamma=C_{P} / C_{V}$. The ratio of ' $\gamma$ ' for monoatomic gas to ' $\gamma$ ' for (rigid) diatomic gas is
1 $\frac{15}{35}$
2 $\frac{35}{15}$
3 $\frac{21}{25}$
4 $\frac{25}{21}$
Explanation:
D $\mathrm{f}=$ number of degrees of freedom $\gamma=\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=1+\frac{2}{\mathrm{f}}$ For monatomic gas $\mathrm{f}_{1}=3$ For diatomic gas $\mathrm{f}_{2}=5$ Ratio of $\gamma$ for monatomic gas to $\gamma$ for diatomic gas $\frac{\gamma_{1}}{\gamma_{2}}=\frac{1+\frac{2}{f_{1}}}{1+\frac{2}{f_{2}}}$ $\frac{\gamma_{1}}{\gamma_{2}}=\frac{1+\frac{2}{3}}{1+\frac{2}{5}}$ $=\frac{\frac{5}{3}}{\frac{7}{5}}=\frac{25}{21}$
MHT-CET 2020
Kinetic Theory of Gases
139125
If a gas has $n$ degrees of freedom, then the ratio of principal specific heats of the gas is
1 $1+\frac{2}{\mathrm{n}}$
2 $1+\frac{n}{2}$
3 $1-\frac{2}{\mathrm{n}}$
4 $1-\frac{n}{2}$
Explanation:
A We know that, $\mathrm{C}_{\mathrm{v}}=\frac{\mathrm{n}}{2} \cdot \mathrm{R}$ And, $C_{p}-C_{v}=R$ $C_{p}=C_{v}+R$ $C_{p}=\frac{n}{2} \cdot R+R$ $\mathrm{C}_{\mathrm{p}}=\mathrm{R}\left(1+\frac{\mathrm{n}}{2}\right)$ Now, ratio of specific heats $\gamma=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}$ $\gamma=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\frac{\mathrm{R}\left(1+\frac{\mathrm{n}}{2}\right)}{\frac{\mathrm{n}}{2} \cdot \mathrm{R}}=\frac{2}{\mathrm{n}}+1$ Specific heat ratio of gas $(\gamma)=\frac{2}{n}+1$
139122
A polyatomic gas has $f$ vibrational degrees of freedom, then the ratio of the specific heat at constant pressure to that at constant volume will be
1 $\frac{4+\mathrm{f}}{3+\mathrm{f}}$
2 $\frac{4-\mathrm{f}}{3-\mathrm{f}}$
3 $\frac{3+\mathrm{f}}{3+\mathrm{f}}$
4 $\frac{3-\mathrm{f}}{4-\mathrm{f}}$
Explanation:
A We know that, $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}$ $\mathrm{f}=$ vibrational degrees of freedom For polyatomic gas the specific heat at constant volume, $\mathrm{C}_{\mathrm{V}}=(3+\mathrm{f}) \mathrm{R}$ So, from equation (i), we get- $C_{P}=C_{V}+R$ $C_{P}=(3+f) R+R$ $C_{P}=(4+f) R$ Ratio of $\frac{C_{P}}{C_{V}}=\frac{(4+f) R}{(3+f) R}=\frac{(4+f)}{(3+f)}$
Shift-1]
Kinetic Theory of Gases
139123
The temperature at which the r.m.s. velocity of oxygen molecules will be same as that of hydrogen molecules at $-173^{\circ} \mathrm{C}$ is
1 $1600^{\circ} \mathrm{C}$
2 $373^{\circ} \mathrm{C}$
3 $1327^{\circ} \mathrm{C}$
4 $1200^{\circ} \mathrm{C}$
Explanation:
C We have, $\mathrm{M}_{\mathrm{O}_{2}}=32$ $\mathrm{M}_{\mathrm{H}_{2}}=2$ Temperature of hydrogen molecules $\mathrm{T}_{1}=273+\left(-173^{\circ}\right)$ $\mathrm{T}_{1}=100 \mathrm{~K}$ Temperature of oxygen molecules $=T_{2}$ $\left(\mathrm{V}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}=\sqrt{\frac{3 \mathrm{RT}_{2}}{\mathrm{M}_{\mathrm{O}_{2}}}}$ $\left(\mathrm{~V}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}=\sqrt{\frac{3 \mathrm{RT}_{1}}{\mathrm{M}_{\mathrm{H}_{2}}}}$ According to question $\left(\mathrm{V}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}=\left(\mathrm{V}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}$ $\sqrt{\frac{3 \mathrm{RT}_{2}}{\mathrm{M}_{\mathrm{O}_{2}}}}=\sqrt{\frac{3 \mathrm{RT}_{1}}{\mathrm{M}_{\mathrm{H}_{2}}}}$ $\frac{3 \times \mathrm{T}_{2} \times \mathrm{R}}{32}=\frac{3 \times 100 \times \mathrm{R}}{2}$ $3 \times \mathrm{T}_{2} \times \mathrm{R}=4800 \times \mathrm{R}$ $\mathrm{T}_{2}=1600 \mathrm{~K}$ $\mathrm{~T}_{2}=(1600-273)^{\circ} \mathrm{C}$ $\mathrm{T}_{2}=1327^{\circ} \mathrm{C}$
COMEDK 2019
Kinetic Theory of Gases
139124
If ' $f$ ' is the number of degrees of freedom of a molecule of a gas and ratio of molar specific heats of a gas, $\gamma=1+\frac{2}{f}$ where $\gamma=C_{P} / C_{V}$. The ratio of ' $\gamma$ ' for monoatomic gas to ' $\gamma$ ' for (rigid) diatomic gas is
1 $\frac{15}{35}$
2 $\frac{35}{15}$
3 $\frac{21}{25}$
4 $\frac{25}{21}$
Explanation:
D $\mathrm{f}=$ number of degrees of freedom $\gamma=\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=1+\frac{2}{\mathrm{f}}$ For monatomic gas $\mathrm{f}_{1}=3$ For diatomic gas $\mathrm{f}_{2}=5$ Ratio of $\gamma$ for monatomic gas to $\gamma$ for diatomic gas $\frac{\gamma_{1}}{\gamma_{2}}=\frac{1+\frac{2}{f_{1}}}{1+\frac{2}{f_{2}}}$ $\frac{\gamma_{1}}{\gamma_{2}}=\frac{1+\frac{2}{3}}{1+\frac{2}{5}}$ $=\frac{\frac{5}{3}}{\frac{7}{5}}=\frac{25}{21}$
MHT-CET 2020
Kinetic Theory of Gases
139125
If a gas has $n$ degrees of freedom, then the ratio of principal specific heats of the gas is
1 $1+\frac{2}{\mathrm{n}}$
2 $1+\frac{n}{2}$
3 $1-\frac{2}{\mathrm{n}}$
4 $1-\frac{n}{2}$
Explanation:
A We know that, $\mathrm{C}_{\mathrm{v}}=\frac{\mathrm{n}}{2} \cdot \mathrm{R}$ And, $C_{p}-C_{v}=R$ $C_{p}=C_{v}+R$ $C_{p}=\frac{n}{2} \cdot R+R$ $\mathrm{C}_{\mathrm{p}}=\mathrm{R}\left(1+\frac{\mathrm{n}}{2}\right)$ Now, ratio of specific heats $\gamma=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}$ $\gamma=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\frac{\mathrm{R}\left(1+\frac{\mathrm{n}}{2}\right)}{\frac{\mathrm{n}}{2} \cdot \mathrm{R}}=\frac{2}{\mathrm{n}}+1$ Specific heat ratio of gas $(\gamma)=\frac{2}{n}+1$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Kinetic Theory of Gases
139122
A polyatomic gas has $f$ vibrational degrees of freedom, then the ratio of the specific heat at constant pressure to that at constant volume will be
1 $\frac{4+\mathrm{f}}{3+\mathrm{f}}$
2 $\frac{4-\mathrm{f}}{3-\mathrm{f}}$
3 $\frac{3+\mathrm{f}}{3+\mathrm{f}}$
4 $\frac{3-\mathrm{f}}{4-\mathrm{f}}$
Explanation:
A We know that, $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}$ $\mathrm{f}=$ vibrational degrees of freedom For polyatomic gas the specific heat at constant volume, $\mathrm{C}_{\mathrm{V}}=(3+\mathrm{f}) \mathrm{R}$ So, from equation (i), we get- $C_{P}=C_{V}+R$ $C_{P}=(3+f) R+R$ $C_{P}=(4+f) R$ Ratio of $\frac{C_{P}}{C_{V}}=\frac{(4+f) R}{(3+f) R}=\frac{(4+f)}{(3+f)}$
Shift-1]
Kinetic Theory of Gases
139123
The temperature at which the r.m.s. velocity of oxygen molecules will be same as that of hydrogen molecules at $-173^{\circ} \mathrm{C}$ is
1 $1600^{\circ} \mathrm{C}$
2 $373^{\circ} \mathrm{C}$
3 $1327^{\circ} \mathrm{C}$
4 $1200^{\circ} \mathrm{C}$
Explanation:
C We have, $\mathrm{M}_{\mathrm{O}_{2}}=32$ $\mathrm{M}_{\mathrm{H}_{2}}=2$ Temperature of hydrogen molecules $\mathrm{T}_{1}=273+\left(-173^{\circ}\right)$ $\mathrm{T}_{1}=100 \mathrm{~K}$ Temperature of oxygen molecules $=T_{2}$ $\left(\mathrm{V}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}=\sqrt{\frac{3 \mathrm{RT}_{2}}{\mathrm{M}_{\mathrm{O}_{2}}}}$ $\left(\mathrm{~V}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}=\sqrt{\frac{3 \mathrm{RT}_{1}}{\mathrm{M}_{\mathrm{H}_{2}}}}$ According to question $\left(\mathrm{V}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}=\left(\mathrm{V}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}$ $\sqrt{\frac{3 \mathrm{RT}_{2}}{\mathrm{M}_{\mathrm{O}_{2}}}}=\sqrt{\frac{3 \mathrm{RT}_{1}}{\mathrm{M}_{\mathrm{H}_{2}}}}$ $\frac{3 \times \mathrm{T}_{2} \times \mathrm{R}}{32}=\frac{3 \times 100 \times \mathrm{R}}{2}$ $3 \times \mathrm{T}_{2} \times \mathrm{R}=4800 \times \mathrm{R}$ $\mathrm{T}_{2}=1600 \mathrm{~K}$ $\mathrm{~T}_{2}=(1600-273)^{\circ} \mathrm{C}$ $\mathrm{T}_{2}=1327^{\circ} \mathrm{C}$
COMEDK 2019
Kinetic Theory of Gases
139124
If ' $f$ ' is the number of degrees of freedom of a molecule of a gas and ratio of molar specific heats of a gas, $\gamma=1+\frac{2}{f}$ where $\gamma=C_{P} / C_{V}$. The ratio of ' $\gamma$ ' for monoatomic gas to ' $\gamma$ ' for (rigid) diatomic gas is
1 $\frac{15}{35}$
2 $\frac{35}{15}$
3 $\frac{21}{25}$
4 $\frac{25}{21}$
Explanation:
D $\mathrm{f}=$ number of degrees of freedom $\gamma=\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=1+\frac{2}{\mathrm{f}}$ For monatomic gas $\mathrm{f}_{1}=3$ For diatomic gas $\mathrm{f}_{2}=5$ Ratio of $\gamma$ for monatomic gas to $\gamma$ for diatomic gas $\frac{\gamma_{1}}{\gamma_{2}}=\frac{1+\frac{2}{f_{1}}}{1+\frac{2}{f_{2}}}$ $\frac{\gamma_{1}}{\gamma_{2}}=\frac{1+\frac{2}{3}}{1+\frac{2}{5}}$ $=\frac{\frac{5}{3}}{\frac{7}{5}}=\frac{25}{21}$
MHT-CET 2020
Kinetic Theory of Gases
139125
If a gas has $n$ degrees of freedom, then the ratio of principal specific heats of the gas is
1 $1+\frac{2}{\mathrm{n}}$
2 $1+\frac{n}{2}$
3 $1-\frac{2}{\mathrm{n}}$
4 $1-\frac{n}{2}$
Explanation:
A We know that, $\mathrm{C}_{\mathrm{v}}=\frac{\mathrm{n}}{2} \cdot \mathrm{R}$ And, $C_{p}-C_{v}=R$ $C_{p}=C_{v}+R$ $C_{p}=\frac{n}{2} \cdot R+R$ $\mathrm{C}_{\mathrm{p}}=\mathrm{R}\left(1+\frac{\mathrm{n}}{2}\right)$ Now, ratio of specific heats $\gamma=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}$ $\gamma=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\frac{\mathrm{R}\left(1+\frac{\mathrm{n}}{2}\right)}{\frac{\mathrm{n}}{2} \cdot \mathrm{R}}=\frac{2}{\mathrm{n}}+1$ Specific heat ratio of gas $(\gamma)=\frac{2}{n}+1$
139122
A polyatomic gas has $f$ vibrational degrees of freedom, then the ratio of the specific heat at constant pressure to that at constant volume will be
1 $\frac{4+\mathrm{f}}{3+\mathrm{f}}$
2 $\frac{4-\mathrm{f}}{3-\mathrm{f}}$
3 $\frac{3+\mathrm{f}}{3+\mathrm{f}}$
4 $\frac{3-\mathrm{f}}{4-\mathrm{f}}$
Explanation:
A We know that, $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}$ $\mathrm{f}=$ vibrational degrees of freedom For polyatomic gas the specific heat at constant volume, $\mathrm{C}_{\mathrm{V}}=(3+\mathrm{f}) \mathrm{R}$ So, from equation (i), we get- $C_{P}=C_{V}+R$ $C_{P}=(3+f) R+R$ $C_{P}=(4+f) R$ Ratio of $\frac{C_{P}}{C_{V}}=\frac{(4+f) R}{(3+f) R}=\frac{(4+f)}{(3+f)}$
Shift-1]
Kinetic Theory of Gases
139123
The temperature at which the r.m.s. velocity of oxygen molecules will be same as that of hydrogen molecules at $-173^{\circ} \mathrm{C}$ is
1 $1600^{\circ} \mathrm{C}$
2 $373^{\circ} \mathrm{C}$
3 $1327^{\circ} \mathrm{C}$
4 $1200^{\circ} \mathrm{C}$
Explanation:
C We have, $\mathrm{M}_{\mathrm{O}_{2}}=32$ $\mathrm{M}_{\mathrm{H}_{2}}=2$ Temperature of hydrogen molecules $\mathrm{T}_{1}=273+\left(-173^{\circ}\right)$ $\mathrm{T}_{1}=100 \mathrm{~K}$ Temperature of oxygen molecules $=T_{2}$ $\left(\mathrm{V}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}=\sqrt{\frac{3 \mathrm{RT}_{2}}{\mathrm{M}_{\mathrm{O}_{2}}}}$ $\left(\mathrm{~V}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}=\sqrt{\frac{3 \mathrm{RT}_{1}}{\mathrm{M}_{\mathrm{H}_{2}}}}$ According to question $\left(\mathrm{V}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}=\left(\mathrm{V}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}$ $\sqrt{\frac{3 \mathrm{RT}_{2}}{\mathrm{M}_{\mathrm{O}_{2}}}}=\sqrt{\frac{3 \mathrm{RT}_{1}}{\mathrm{M}_{\mathrm{H}_{2}}}}$ $\frac{3 \times \mathrm{T}_{2} \times \mathrm{R}}{32}=\frac{3 \times 100 \times \mathrm{R}}{2}$ $3 \times \mathrm{T}_{2} \times \mathrm{R}=4800 \times \mathrm{R}$ $\mathrm{T}_{2}=1600 \mathrm{~K}$ $\mathrm{~T}_{2}=(1600-273)^{\circ} \mathrm{C}$ $\mathrm{T}_{2}=1327^{\circ} \mathrm{C}$
COMEDK 2019
Kinetic Theory of Gases
139124
If ' $f$ ' is the number of degrees of freedom of a molecule of a gas and ratio of molar specific heats of a gas, $\gamma=1+\frac{2}{f}$ where $\gamma=C_{P} / C_{V}$. The ratio of ' $\gamma$ ' for monoatomic gas to ' $\gamma$ ' for (rigid) diatomic gas is
1 $\frac{15}{35}$
2 $\frac{35}{15}$
3 $\frac{21}{25}$
4 $\frac{25}{21}$
Explanation:
D $\mathrm{f}=$ number of degrees of freedom $\gamma=\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=1+\frac{2}{\mathrm{f}}$ For monatomic gas $\mathrm{f}_{1}=3$ For diatomic gas $\mathrm{f}_{2}=5$ Ratio of $\gamma$ for monatomic gas to $\gamma$ for diatomic gas $\frac{\gamma_{1}}{\gamma_{2}}=\frac{1+\frac{2}{f_{1}}}{1+\frac{2}{f_{2}}}$ $\frac{\gamma_{1}}{\gamma_{2}}=\frac{1+\frac{2}{3}}{1+\frac{2}{5}}$ $=\frac{\frac{5}{3}}{\frac{7}{5}}=\frac{25}{21}$
MHT-CET 2020
Kinetic Theory of Gases
139125
If a gas has $n$ degrees of freedom, then the ratio of principal specific heats of the gas is
1 $1+\frac{2}{\mathrm{n}}$
2 $1+\frac{n}{2}$
3 $1-\frac{2}{\mathrm{n}}$
4 $1-\frac{n}{2}$
Explanation:
A We know that, $\mathrm{C}_{\mathrm{v}}=\frac{\mathrm{n}}{2} \cdot \mathrm{R}$ And, $C_{p}-C_{v}=R$ $C_{p}=C_{v}+R$ $C_{p}=\frac{n}{2} \cdot R+R$ $\mathrm{C}_{\mathrm{p}}=\mathrm{R}\left(1+\frac{\mathrm{n}}{2}\right)$ Now, ratio of specific heats $\gamma=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}$ $\gamma=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\frac{\mathrm{R}\left(1+\frac{\mathrm{n}}{2}\right)}{\frac{\mathrm{n}}{2} \cdot \mathrm{R}}=\frac{2}{\mathrm{n}}+1$ Specific heat ratio of gas $(\gamma)=\frac{2}{n}+1$