139339
For a monoatomic gas, the molar specific heat at constant pressure divided by the molar gas constant $R$ is equal to
1 2.5
2 1.5
3 5.0
4 3.5
5 4.0
Explanation:
A $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R} \quad$.....(i) For monoatomic gas $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=1.67$ $\mathrm{C}_{\mathrm{P}}=1.67 \mathrm{C}_{\mathrm{V}}$ Multiplying equation (i) by 1.67, we have $1.67 \mathrm{C}_{\mathrm{P}}-1.67 \mathrm{C}_{\mathrm{V}}=1.67 \mathrm{R}$ Putting $1.67 \mathrm{C}_{\mathrm{V}}=\mathrm{C}_{\mathrm{P}}$ in equation (ii), $1.67 \mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{P}}=1.67 \mathrm{R}$ $0.67 \mathrm{C}_{\mathrm{P}}=1.67 \mathrm{R}$ $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{R}}=\frac{1.67}{0.67}=2.49$ $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{R}} \cong 2.5$
Kerala CEE - 2015
Kinetic Theory of Gases
139331
For a gas the value of $\frac{R}{C_{v}}=0.4$, so the gas is (R-universal gas constant) #[Qdiff: Hard, QCat: Numerical Based, examname: AP EAMCET (21.04.2019) Shift-II#
1 monoatomic
2 diatomic
3 triatomic
4 polyatomic
Explanation:
B $\frac{\mathrm{R}}{\mathrm{C}_{\mathrm{V}}}=0.4$ (Given) We know that, $\gamma-1=\frac{R}{C_{V}}$ $\gamma-1=0.4$ $\gamma=1.4$ $\left(\because \frac{\mathrm{R}}{\mathrm{C}_{\mathrm{v}}}=0.4\right)$ $\gamma=\frac{14}{10}=\frac{7}{5}$ $\gamma=\frac{7}{5}$ For diatomic gas $\gamma=\frac{7}{5}$
Kinetic Theory of Gases
139341
Two moles of oxygen is mixed with eight moles of helium. The effective specific heat of the mixture at constant volume is
1 $1.3 \mathrm{R}$
2 $1.4 \mathrm{R}$
3 $1.7 \mathrm{R}$
4 $1.9 \mathrm{R}$
5 $1.2 \mathrm{R}$
Explanation:
C Oxygen $\left(\mathrm{O}_{2}\right)$ is diatomic gas $\left(\mathrm{C}_{\mathrm{v}}\right)_{\mathrm{O}_{2}}=\frac{5 \mathrm{R}}{2}$ Helium is monoatomic gas So, $\left(\mathrm{C}_{\mathrm{v}}\right)_{\mathrm{He}}=\frac{3 \mathrm{R}}{2}$ Moles of Oxygen $n_{1}=2$ Moles of Helium $\mathrm{n}_{2}=8$ Let, effective specific heat of the mixture at constant volume is $\mathrm{C}_{\mathrm{V}}$ Then, $\mathrm{C}_{\mathrm{V}}=\frac{\mathrm{n}_{1}\left(\mathrm{C}_{\mathrm{v}}\right)_{\mathrm{O}_{2}}+\mathrm{n}_{2}\left(\mathrm{C}_{\mathrm{v}}\right)_{\mathrm{He}}}{\mathrm{n}_{1}+\mathrm{n}_{2}}$ $\mathrm{C}_{\mathrm{V}}= \frac{2 \times \frac{5 \mathrm{R}}{2}+8 \times \frac{3 \mathrm{R}}{2}}{2+8}$ $= \frac{5 \mathrm{R}+12 \mathrm{R}}{10}=1.7 \mathrm{R}$ $\mathrm{C}_{\mathrm{V}}=1.7 \mathrm{R}$
Kerala CEE 2007
Kinetic Theory of Gases
139342
One mole of a gas occupies 22.4 lit at N.T.P. Calculate the difference between two molar specific heats of the gas. $\mathrm{J}=\mathbf{4 2 0 0} \mathrm{J} / \mathrm{kcal}$.
139339
For a monoatomic gas, the molar specific heat at constant pressure divided by the molar gas constant $R$ is equal to
1 2.5
2 1.5
3 5.0
4 3.5
5 4.0
Explanation:
A $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R} \quad$.....(i) For monoatomic gas $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=1.67$ $\mathrm{C}_{\mathrm{P}}=1.67 \mathrm{C}_{\mathrm{V}}$ Multiplying equation (i) by 1.67, we have $1.67 \mathrm{C}_{\mathrm{P}}-1.67 \mathrm{C}_{\mathrm{V}}=1.67 \mathrm{R}$ Putting $1.67 \mathrm{C}_{\mathrm{V}}=\mathrm{C}_{\mathrm{P}}$ in equation (ii), $1.67 \mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{P}}=1.67 \mathrm{R}$ $0.67 \mathrm{C}_{\mathrm{P}}=1.67 \mathrm{R}$ $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{R}}=\frac{1.67}{0.67}=2.49$ $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{R}} \cong 2.5$
Kerala CEE - 2015
Kinetic Theory of Gases
139331
For a gas the value of $\frac{R}{C_{v}}=0.4$, so the gas is (R-universal gas constant) #[Qdiff: Hard, QCat: Numerical Based, examname: AP EAMCET (21.04.2019) Shift-II#
1 monoatomic
2 diatomic
3 triatomic
4 polyatomic
Explanation:
B $\frac{\mathrm{R}}{\mathrm{C}_{\mathrm{V}}}=0.4$ (Given) We know that, $\gamma-1=\frac{R}{C_{V}}$ $\gamma-1=0.4$ $\gamma=1.4$ $\left(\because \frac{\mathrm{R}}{\mathrm{C}_{\mathrm{v}}}=0.4\right)$ $\gamma=\frac{14}{10}=\frac{7}{5}$ $\gamma=\frac{7}{5}$ For diatomic gas $\gamma=\frac{7}{5}$
Kinetic Theory of Gases
139341
Two moles of oxygen is mixed with eight moles of helium. The effective specific heat of the mixture at constant volume is
1 $1.3 \mathrm{R}$
2 $1.4 \mathrm{R}$
3 $1.7 \mathrm{R}$
4 $1.9 \mathrm{R}$
5 $1.2 \mathrm{R}$
Explanation:
C Oxygen $\left(\mathrm{O}_{2}\right)$ is diatomic gas $\left(\mathrm{C}_{\mathrm{v}}\right)_{\mathrm{O}_{2}}=\frac{5 \mathrm{R}}{2}$ Helium is monoatomic gas So, $\left(\mathrm{C}_{\mathrm{v}}\right)_{\mathrm{He}}=\frac{3 \mathrm{R}}{2}$ Moles of Oxygen $n_{1}=2$ Moles of Helium $\mathrm{n}_{2}=8$ Let, effective specific heat of the mixture at constant volume is $\mathrm{C}_{\mathrm{V}}$ Then, $\mathrm{C}_{\mathrm{V}}=\frac{\mathrm{n}_{1}\left(\mathrm{C}_{\mathrm{v}}\right)_{\mathrm{O}_{2}}+\mathrm{n}_{2}\left(\mathrm{C}_{\mathrm{v}}\right)_{\mathrm{He}}}{\mathrm{n}_{1}+\mathrm{n}_{2}}$ $\mathrm{C}_{\mathrm{V}}= \frac{2 \times \frac{5 \mathrm{R}}{2}+8 \times \frac{3 \mathrm{R}}{2}}{2+8}$ $= \frac{5 \mathrm{R}+12 \mathrm{R}}{10}=1.7 \mathrm{R}$ $\mathrm{C}_{\mathrm{V}}=1.7 \mathrm{R}$
Kerala CEE 2007
Kinetic Theory of Gases
139342
One mole of a gas occupies 22.4 lit at N.T.P. Calculate the difference between two molar specific heats of the gas. $\mathrm{J}=\mathbf{4 2 0 0} \mathrm{J} / \mathrm{kcal}$.
139339
For a monoatomic gas, the molar specific heat at constant pressure divided by the molar gas constant $R$ is equal to
1 2.5
2 1.5
3 5.0
4 3.5
5 4.0
Explanation:
A $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R} \quad$.....(i) For monoatomic gas $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=1.67$ $\mathrm{C}_{\mathrm{P}}=1.67 \mathrm{C}_{\mathrm{V}}$ Multiplying equation (i) by 1.67, we have $1.67 \mathrm{C}_{\mathrm{P}}-1.67 \mathrm{C}_{\mathrm{V}}=1.67 \mathrm{R}$ Putting $1.67 \mathrm{C}_{\mathrm{V}}=\mathrm{C}_{\mathrm{P}}$ in equation (ii), $1.67 \mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{P}}=1.67 \mathrm{R}$ $0.67 \mathrm{C}_{\mathrm{P}}=1.67 \mathrm{R}$ $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{R}}=\frac{1.67}{0.67}=2.49$ $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{R}} \cong 2.5$
Kerala CEE - 2015
Kinetic Theory of Gases
139331
For a gas the value of $\frac{R}{C_{v}}=0.4$, so the gas is (R-universal gas constant) #[Qdiff: Hard, QCat: Numerical Based, examname: AP EAMCET (21.04.2019) Shift-II#
1 monoatomic
2 diatomic
3 triatomic
4 polyatomic
Explanation:
B $\frac{\mathrm{R}}{\mathrm{C}_{\mathrm{V}}}=0.4$ (Given) We know that, $\gamma-1=\frac{R}{C_{V}}$ $\gamma-1=0.4$ $\gamma=1.4$ $\left(\because \frac{\mathrm{R}}{\mathrm{C}_{\mathrm{v}}}=0.4\right)$ $\gamma=\frac{14}{10}=\frac{7}{5}$ $\gamma=\frac{7}{5}$ For diatomic gas $\gamma=\frac{7}{5}$
Kinetic Theory of Gases
139341
Two moles of oxygen is mixed with eight moles of helium. The effective specific heat of the mixture at constant volume is
1 $1.3 \mathrm{R}$
2 $1.4 \mathrm{R}$
3 $1.7 \mathrm{R}$
4 $1.9 \mathrm{R}$
5 $1.2 \mathrm{R}$
Explanation:
C Oxygen $\left(\mathrm{O}_{2}\right)$ is diatomic gas $\left(\mathrm{C}_{\mathrm{v}}\right)_{\mathrm{O}_{2}}=\frac{5 \mathrm{R}}{2}$ Helium is monoatomic gas So, $\left(\mathrm{C}_{\mathrm{v}}\right)_{\mathrm{He}}=\frac{3 \mathrm{R}}{2}$ Moles of Oxygen $n_{1}=2$ Moles of Helium $\mathrm{n}_{2}=8$ Let, effective specific heat of the mixture at constant volume is $\mathrm{C}_{\mathrm{V}}$ Then, $\mathrm{C}_{\mathrm{V}}=\frac{\mathrm{n}_{1}\left(\mathrm{C}_{\mathrm{v}}\right)_{\mathrm{O}_{2}}+\mathrm{n}_{2}\left(\mathrm{C}_{\mathrm{v}}\right)_{\mathrm{He}}}{\mathrm{n}_{1}+\mathrm{n}_{2}}$ $\mathrm{C}_{\mathrm{V}}= \frac{2 \times \frac{5 \mathrm{R}}{2}+8 \times \frac{3 \mathrm{R}}{2}}{2+8}$ $= \frac{5 \mathrm{R}+12 \mathrm{R}}{10}=1.7 \mathrm{R}$ $\mathrm{C}_{\mathrm{V}}=1.7 \mathrm{R}$
Kerala CEE 2007
Kinetic Theory of Gases
139342
One mole of a gas occupies 22.4 lit at N.T.P. Calculate the difference between two molar specific heats of the gas. $\mathrm{J}=\mathbf{4 2 0 0} \mathrm{J} / \mathrm{kcal}$.
139339
For a monoatomic gas, the molar specific heat at constant pressure divided by the molar gas constant $R$ is equal to
1 2.5
2 1.5
3 5.0
4 3.5
5 4.0
Explanation:
A $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R} \quad$.....(i) For monoatomic gas $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=1.67$ $\mathrm{C}_{\mathrm{P}}=1.67 \mathrm{C}_{\mathrm{V}}$ Multiplying equation (i) by 1.67, we have $1.67 \mathrm{C}_{\mathrm{P}}-1.67 \mathrm{C}_{\mathrm{V}}=1.67 \mathrm{R}$ Putting $1.67 \mathrm{C}_{\mathrm{V}}=\mathrm{C}_{\mathrm{P}}$ in equation (ii), $1.67 \mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{P}}=1.67 \mathrm{R}$ $0.67 \mathrm{C}_{\mathrm{P}}=1.67 \mathrm{R}$ $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{R}}=\frac{1.67}{0.67}=2.49$ $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{R}} \cong 2.5$
Kerala CEE - 2015
Kinetic Theory of Gases
139331
For a gas the value of $\frac{R}{C_{v}}=0.4$, so the gas is (R-universal gas constant) #[Qdiff: Hard, QCat: Numerical Based, examname: AP EAMCET (21.04.2019) Shift-II#
1 monoatomic
2 diatomic
3 triatomic
4 polyatomic
Explanation:
B $\frac{\mathrm{R}}{\mathrm{C}_{\mathrm{V}}}=0.4$ (Given) We know that, $\gamma-1=\frac{R}{C_{V}}$ $\gamma-1=0.4$ $\gamma=1.4$ $\left(\because \frac{\mathrm{R}}{\mathrm{C}_{\mathrm{v}}}=0.4\right)$ $\gamma=\frac{14}{10}=\frac{7}{5}$ $\gamma=\frac{7}{5}$ For diatomic gas $\gamma=\frac{7}{5}$
Kinetic Theory of Gases
139341
Two moles of oxygen is mixed with eight moles of helium. The effective specific heat of the mixture at constant volume is
1 $1.3 \mathrm{R}$
2 $1.4 \mathrm{R}$
3 $1.7 \mathrm{R}$
4 $1.9 \mathrm{R}$
5 $1.2 \mathrm{R}$
Explanation:
C Oxygen $\left(\mathrm{O}_{2}\right)$ is diatomic gas $\left(\mathrm{C}_{\mathrm{v}}\right)_{\mathrm{O}_{2}}=\frac{5 \mathrm{R}}{2}$ Helium is monoatomic gas So, $\left(\mathrm{C}_{\mathrm{v}}\right)_{\mathrm{He}}=\frac{3 \mathrm{R}}{2}$ Moles of Oxygen $n_{1}=2$ Moles of Helium $\mathrm{n}_{2}=8$ Let, effective specific heat of the mixture at constant volume is $\mathrm{C}_{\mathrm{V}}$ Then, $\mathrm{C}_{\mathrm{V}}=\frac{\mathrm{n}_{1}\left(\mathrm{C}_{\mathrm{v}}\right)_{\mathrm{O}_{2}}+\mathrm{n}_{2}\left(\mathrm{C}_{\mathrm{v}}\right)_{\mathrm{He}}}{\mathrm{n}_{1}+\mathrm{n}_{2}}$ $\mathrm{C}_{\mathrm{V}}= \frac{2 \times \frac{5 \mathrm{R}}{2}+8 \times \frac{3 \mathrm{R}}{2}}{2+8}$ $= \frac{5 \mathrm{R}+12 \mathrm{R}}{10}=1.7 \mathrm{R}$ $\mathrm{C}_{\mathrm{V}}=1.7 \mathrm{R}$
Kerala CEE 2007
Kinetic Theory of Gases
139342
One mole of a gas occupies 22.4 lit at N.T.P. Calculate the difference between two molar specific heats of the gas. $\mathrm{J}=\mathbf{4 2 0 0} \mathrm{J} / \mathrm{kcal}$.
