139337
An ideal gas has pressure ' $P$ ', volume ' $V$ ' and absolute temperature ' $T$ '. If ' $m$ ' is the mass of each molecule and ' $K$ ' is the Boltzmann constant then density of the gas is
1 $\frac{\mathrm{Pm}}{\mathrm{KT}}$
2 $\frac{\mathrm{KT}}{\mathrm{Pm}}$
3 $\frac{\mathrm{Km}}{\mathrm{PT}}$
4 $\frac{\mathrm{PK}}{\mathrm{Tm}}$
Explanation:
A Given that, $\mathrm{m}=$ mass of each molecule $\mathrm{K}=$ Boltzmann constant $\mathrm{N}=$ Number of molecules $\mathrm{N}_{\mathrm{A}}=$ Avogadro number Using general gas equation, $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{n}=$ no. of moles $=\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{A}}}$ Total mass $=\mathrm{mN}$ Density $(\rho)=\frac{\mathrm{mN}}{\mathrm{V}}$ $\mathrm{V}=\frac{\mathrm{mN}}{\rho}$ Putting respective values in equation (i), $\mathrm{P}\left(\frac{\mathrm{mN}}{\rho}\right)=\left(\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{A}}}\right)\left(\mathrm{KN}_{\mathrm{A}}\right) \mathrm{T}$ $\frac{\mathrm{Pm}}{\rho}=\mathrm{KT}$ $\rho=\frac{\text { Pm }}{\text { KT }}$
MHT-CET 2020
Kinetic Theory of Gases
139340
The ratio of the molar heat capacities of a diatomic gas at constant pressure to that at constant volume is
1 $\frac{7}{2}$
2 $\frac{3}{2}$
3 $\frac{3}{5}$
4 $\frac{7}{5}$
5 $\frac{5}{2}$
Explanation:
D For diatomic gas $\mathrm{C}_{\mathrm{P}}=\frac{7}{2} \mathrm{R}, \mathrm{C}_{\mathrm{V}}=\frac{5}{2} \mathrm{R}$ $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma=\frac{\frac{7}{2} \mathrm{R}}{\frac{5}{2} \mathrm{R}}=\frac{7}{5}$ $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma=\frac{7}{5}$
Kerala CEE - 2011
Kinetic Theory of Gases
139349
The ratio of $\frac{C_{p}}{C_{v}}$ for a diatomic gas is
1 $\frac{5}{7}$
2 $\frac{7}{9}$
3 $\frac{5}{3}$
4 $\frac{7}{5}$
Explanation:
D For diatomic gas specific heat at constant volume $C_{V}$ and at constant pressure $C_{P}$. $\mathrm{C}_{\mathrm{V}}=\frac{5}{2} \mathrm{R}$ and $\mathrm{C}_{\mathrm{P}}=\mathrm{R}+\mathrm{C}_{\mathrm{V}}=\frac{5}{2} \mathrm{R}+\mathrm{R}=\frac{7}{2} \mathrm{R}$ Ratio $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{\frac{7}{2} \mathrm{R}}{\frac{5}{2} \mathrm{R}}=\frac{7}{5}$
COMEDK 2014
Kinetic Theory of Gases
139350
The average kinetic energy of a gas molecule is
1 proportional to pressure of gas
2 inversely proportional to volume of gas
3 inversely proportional to absolute temperature of gas
4 proportional to absolute temperature of gas
Explanation:
D The average kinetic energy of gas molecule is given as, $\mathrm{K} . \mathrm{E}=\frac{3}{2} \mathrm{KT}$ $\mathrm{K} . \mathrm{E} \propto \mathrm{T}$ Therefore, average kinetic energy is proportional to absolute temperature.
139337
An ideal gas has pressure ' $P$ ', volume ' $V$ ' and absolute temperature ' $T$ '. If ' $m$ ' is the mass of each molecule and ' $K$ ' is the Boltzmann constant then density of the gas is
1 $\frac{\mathrm{Pm}}{\mathrm{KT}}$
2 $\frac{\mathrm{KT}}{\mathrm{Pm}}$
3 $\frac{\mathrm{Km}}{\mathrm{PT}}$
4 $\frac{\mathrm{PK}}{\mathrm{Tm}}$
Explanation:
A Given that, $\mathrm{m}=$ mass of each molecule $\mathrm{K}=$ Boltzmann constant $\mathrm{N}=$ Number of molecules $\mathrm{N}_{\mathrm{A}}=$ Avogadro number Using general gas equation, $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{n}=$ no. of moles $=\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{A}}}$ Total mass $=\mathrm{mN}$ Density $(\rho)=\frac{\mathrm{mN}}{\mathrm{V}}$ $\mathrm{V}=\frac{\mathrm{mN}}{\rho}$ Putting respective values in equation (i), $\mathrm{P}\left(\frac{\mathrm{mN}}{\rho}\right)=\left(\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{A}}}\right)\left(\mathrm{KN}_{\mathrm{A}}\right) \mathrm{T}$ $\frac{\mathrm{Pm}}{\rho}=\mathrm{KT}$ $\rho=\frac{\text { Pm }}{\text { KT }}$
MHT-CET 2020
Kinetic Theory of Gases
139340
The ratio of the molar heat capacities of a diatomic gas at constant pressure to that at constant volume is
1 $\frac{7}{2}$
2 $\frac{3}{2}$
3 $\frac{3}{5}$
4 $\frac{7}{5}$
5 $\frac{5}{2}$
Explanation:
D For diatomic gas $\mathrm{C}_{\mathrm{P}}=\frac{7}{2} \mathrm{R}, \mathrm{C}_{\mathrm{V}}=\frac{5}{2} \mathrm{R}$ $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma=\frac{\frac{7}{2} \mathrm{R}}{\frac{5}{2} \mathrm{R}}=\frac{7}{5}$ $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma=\frac{7}{5}$
Kerala CEE - 2011
Kinetic Theory of Gases
139349
The ratio of $\frac{C_{p}}{C_{v}}$ for a diatomic gas is
1 $\frac{5}{7}$
2 $\frac{7}{9}$
3 $\frac{5}{3}$
4 $\frac{7}{5}$
Explanation:
D For diatomic gas specific heat at constant volume $C_{V}$ and at constant pressure $C_{P}$. $\mathrm{C}_{\mathrm{V}}=\frac{5}{2} \mathrm{R}$ and $\mathrm{C}_{\mathrm{P}}=\mathrm{R}+\mathrm{C}_{\mathrm{V}}=\frac{5}{2} \mathrm{R}+\mathrm{R}=\frac{7}{2} \mathrm{R}$ Ratio $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{\frac{7}{2} \mathrm{R}}{\frac{5}{2} \mathrm{R}}=\frac{7}{5}$
COMEDK 2014
Kinetic Theory of Gases
139350
The average kinetic energy of a gas molecule is
1 proportional to pressure of gas
2 inversely proportional to volume of gas
3 inversely proportional to absolute temperature of gas
4 proportional to absolute temperature of gas
Explanation:
D The average kinetic energy of gas molecule is given as, $\mathrm{K} . \mathrm{E}=\frac{3}{2} \mathrm{KT}$ $\mathrm{K} . \mathrm{E} \propto \mathrm{T}$ Therefore, average kinetic energy is proportional to absolute temperature.
NEET Test Series from KOTA - 10 Papers In MS WORD
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Kinetic Theory of Gases
139337
An ideal gas has pressure ' $P$ ', volume ' $V$ ' and absolute temperature ' $T$ '. If ' $m$ ' is the mass of each molecule and ' $K$ ' is the Boltzmann constant then density of the gas is
1 $\frac{\mathrm{Pm}}{\mathrm{KT}}$
2 $\frac{\mathrm{KT}}{\mathrm{Pm}}$
3 $\frac{\mathrm{Km}}{\mathrm{PT}}$
4 $\frac{\mathrm{PK}}{\mathrm{Tm}}$
Explanation:
A Given that, $\mathrm{m}=$ mass of each molecule $\mathrm{K}=$ Boltzmann constant $\mathrm{N}=$ Number of molecules $\mathrm{N}_{\mathrm{A}}=$ Avogadro number Using general gas equation, $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{n}=$ no. of moles $=\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{A}}}$ Total mass $=\mathrm{mN}$ Density $(\rho)=\frac{\mathrm{mN}}{\mathrm{V}}$ $\mathrm{V}=\frac{\mathrm{mN}}{\rho}$ Putting respective values in equation (i), $\mathrm{P}\left(\frac{\mathrm{mN}}{\rho}\right)=\left(\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{A}}}\right)\left(\mathrm{KN}_{\mathrm{A}}\right) \mathrm{T}$ $\frac{\mathrm{Pm}}{\rho}=\mathrm{KT}$ $\rho=\frac{\text { Pm }}{\text { KT }}$
MHT-CET 2020
Kinetic Theory of Gases
139340
The ratio of the molar heat capacities of a diatomic gas at constant pressure to that at constant volume is
1 $\frac{7}{2}$
2 $\frac{3}{2}$
3 $\frac{3}{5}$
4 $\frac{7}{5}$
5 $\frac{5}{2}$
Explanation:
D For diatomic gas $\mathrm{C}_{\mathrm{P}}=\frac{7}{2} \mathrm{R}, \mathrm{C}_{\mathrm{V}}=\frac{5}{2} \mathrm{R}$ $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma=\frac{\frac{7}{2} \mathrm{R}}{\frac{5}{2} \mathrm{R}}=\frac{7}{5}$ $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma=\frac{7}{5}$
Kerala CEE - 2011
Kinetic Theory of Gases
139349
The ratio of $\frac{C_{p}}{C_{v}}$ for a diatomic gas is
1 $\frac{5}{7}$
2 $\frac{7}{9}$
3 $\frac{5}{3}$
4 $\frac{7}{5}$
Explanation:
D For diatomic gas specific heat at constant volume $C_{V}$ and at constant pressure $C_{P}$. $\mathrm{C}_{\mathrm{V}}=\frac{5}{2} \mathrm{R}$ and $\mathrm{C}_{\mathrm{P}}=\mathrm{R}+\mathrm{C}_{\mathrm{V}}=\frac{5}{2} \mathrm{R}+\mathrm{R}=\frac{7}{2} \mathrm{R}$ Ratio $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{\frac{7}{2} \mathrm{R}}{\frac{5}{2} \mathrm{R}}=\frac{7}{5}$
COMEDK 2014
Kinetic Theory of Gases
139350
The average kinetic energy of a gas molecule is
1 proportional to pressure of gas
2 inversely proportional to volume of gas
3 inversely proportional to absolute temperature of gas
4 proportional to absolute temperature of gas
Explanation:
D The average kinetic energy of gas molecule is given as, $\mathrm{K} . \mathrm{E}=\frac{3}{2} \mathrm{KT}$ $\mathrm{K} . \mathrm{E} \propto \mathrm{T}$ Therefore, average kinetic energy is proportional to absolute temperature.
139337
An ideal gas has pressure ' $P$ ', volume ' $V$ ' and absolute temperature ' $T$ '. If ' $m$ ' is the mass of each molecule and ' $K$ ' is the Boltzmann constant then density of the gas is
1 $\frac{\mathrm{Pm}}{\mathrm{KT}}$
2 $\frac{\mathrm{KT}}{\mathrm{Pm}}$
3 $\frac{\mathrm{Km}}{\mathrm{PT}}$
4 $\frac{\mathrm{PK}}{\mathrm{Tm}}$
Explanation:
A Given that, $\mathrm{m}=$ mass of each molecule $\mathrm{K}=$ Boltzmann constant $\mathrm{N}=$ Number of molecules $\mathrm{N}_{\mathrm{A}}=$ Avogadro number Using general gas equation, $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{n}=$ no. of moles $=\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{A}}}$ Total mass $=\mathrm{mN}$ Density $(\rho)=\frac{\mathrm{mN}}{\mathrm{V}}$ $\mathrm{V}=\frac{\mathrm{mN}}{\rho}$ Putting respective values in equation (i), $\mathrm{P}\left(\frac{\mathrm{mN}}{\rho}\right)=\left(\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{A}}}\right)\left(\mathrm{KN}_{\mathrm{A}}\right) \mathrm{T}$ $\frac{\mathrm{Pm}}{\rho}=\mathrm{KT}$ $\rho=\frac{\text { Pm }}{\text { KT }}$
MHT-CET 2020
Kinetic Theory of Gases
139340
The ratio of the molar heat capacities of a diatomic gas at constant pressure to that at constant volume is
1 $\frac{7}{2}$
2 $\frac{3}{2}$
3 $\frac{3}{5}$
4 $\frac{7}{5}$
5 $\frac{5}{2}$
Explanation:
D For diatomic gas $\mathrm{C}_{\mathrm{P}}=\frac{7}{2} \mathrm{R}, \mathrm{C}_{\mathrm{V}}=\frac{5}{2} \mathrm{R}$ $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma=\frac{\frac{7}{2} \mathrm{R}}{\frac{5}{2} \mathrm{R}}=\frac{7}{5}$ $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma=\frac{7}{5}$
Kerala CEE - 2011
Kinetic Theory of Gases
139349
The ratio of $\frac{C_{p}}{C_{v}}$ for a diatomic gas is
1 $\frac{5}{7}$
2 $\frac{7}{9}$
3 $\frac{5}{3}$
4 $\frac{7}{5}$
Explanation:
D For diatomic gas specific heat at constant volume $C_{V}$ and at constant pressure $C_{P}$. $\mathrm{C}_{\mathrm{V}}=\frac{5}{2} \mathrm{R}$ and $\mathrm{C}_{\mathrm{P}}=\mathrm{R}+\mathrm{C}_{\mathrm{V}}=\frac{5}{2} \mathrm{R}+\mathrm{R}=\frac{7}{2} \mathrm{R}$ Ratio $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{\frac{7}{2} \mathrm{R}}{\frac{5}{2} \mathrm{R}}=\frac{7}{5}$
COMEDK 2014
Kinetic Theory of Gases
139350
The average kinetic energy of a gas molecule is
1 proportional to pressure of gas
2 inversely proportional to volume of gas
3 inversely proportional to absolute temperature of gas
4 proportional to absolute temperature of gas
Explanation:
D The average kinetic energy of gas molecule is given as, $\mathrm{K} . \mathrm{E}=\frac{3}{2} \mathrm{KT}$ $\mathrm{K} . \mathrm{E} \propto \mathrm{T}$ Therefore, average kinetic energy is proportional to absolute temperature.