139095
The mass of oxygen gas occupying a volume of 11.2 $\mathrm{L}$ at a temperature $27^{\circ} \mathrm{C}$ and a pressure of $760 \mathrm{~nm}$ of mercury in kilogram is [Molecular weight of oxygen $=32$ ]
139096
The tyre of a motor car contains air at $15^{\circ} \mathrm{C}$. If the temperature increases to $35^{\circ} \mathrm{C}$, the approximate percentage increase in pressure is (ignore to expansion of tyre)
138991
An ideal gas equation can be written as $p=\frac{\rho R T}{M_{0}}$ where, $\rho$ and $M_{0}$ are respectively,
1 mass density, mass of the gas
2 number density, molar mass
3 mass density, molar mass
4 number density, mass of the gas
Explanation:
C According to an ideal gas equation- $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{PV}=\frac{\mathrm{m}}{\mathrm{M}_{\mathrm{o}}} \mathrm{RT} \quad\left(\because \mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}_{\mathrm{o}}}\right)$ $\mathrm{P}=\left(\frac{\mathrm{m}}{\mathrm{V}}\right) \frac{\mathrm{RT}}{\mathrm{M}_{\mathrm{o}}}$ $\mathrm{P}=\rho \frac{\mathrm{RT}}{\mathrm{M}_{\mathrm{o}}} \quad\left(\because \rho=\frac{\mathrm{m}}{\mathrm{V}}\right)$ Where $\quad \rho=$ mass density $\mathrm{M}_{\mathrm{o}}=$ Molar mass
NEET OCT-2020
Kinetic Theory of Gases
139028
Pressure of an ideal gas is increased by keeping temperature constant. What is the effect on kinetic energy of molecules?
1 Increase
2 Decrease
3 No change
4 Can't be determined
Explanation:
C Kinetic energy of an ideal gas - $\text { K.E. }=\frac{3}{2} \mathrm{NK}_{\mathrm{B}} \mathrm{T}$ Where, $\mathrm{N}=$ Number of molecules $\mathrm{K}_{\mathrm{B}}=\text { Boltzmann's constant }$ Since kinetic energy is independent of pressure, therefore on increasing the pressure by keeping the temperature constant the kinetic energy of ideal gas remains constant. That means there will be no change on the kinetic energy of molecules.
139095
The mass of oxygen gas occupying a volume of 11.2 $\mathrm{L}$ at a temperature $27^{\circ} \mathrm{C}$ and a pressure of $760 \mathrm{~nm}$ of mercury in kilogram is [Molecular weight of oxygen $=32$ ]
139096
The tyre of a motor car contains air at $15^{\circ} \mathrm{C}$. If the temperature increases to $35^{\circ} \mathrm{C}$, the approximate percentage increase in pressure is (ignore to expansion of tyre)
138991
An ideal gas equation can be written as $p=\frac{\rho R T}{M_{0}}$ where, $\rho$ and $M_{0}$ are respectively,
1 mass density, mass of the gas
2 number density, molar mass
3 mass density, molar mass
4 number density, mass of the gas
Explanation:
C According to an ideal gas equation- $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{PV}=\frac{\mathrm{m}}{\mathrm{M}_{\mathrm{o}}} \mathrm{RT} \quad\left(\because \mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}_{\mathrm{o}}}\right)$ $\mathrm{P}=\left(\frac{\mathrm{m}}{\mathrm{V}}\right) \frac{\mathrm{RT}}{\mathrm{M}_{\mathrm{o}}}$ $\mathrm{P}=\rho \frac{\mathrm{RT}}{\mathrm{M}_{\mathrm{o}}} \quad\left(\because \rho=\frac{\mathrm{m}}{\mathrm{V}}\right)$ Where $\quad \rho=$ mass density $\mathrm{M}_{\mathrm{o}}=$ Molar mass
NEET OCT-2020
Kinetic Theory of Gases
139028
Pressure of an ideal gas is increased by keeping temperature constant. What is the effect on kinetic energy of molecules?
1 Increase
2 Decrease
3 No change
4 Can't be determined
Explanation:
C Kinetic energy of an ideal gas - $\text { K.E. }=\frac{3}{2} \mathrm{NK}_{\mathrm{B}} \mathrm{T}$ Where, $\mathrm{N}=$ Number of molecules $\mathrm{K}_{\mathrm{B}}=\text { Boltzmann's constant }$ Since kinetic energy is independent of pressure, therefore on increasing the pressure by keeping the temperature constant the kinetic energy of ideal gas remains constant. That means there will be no change on the kinetic energy of molecules.
139095
The mass of oxygen gas occupying a volume of 11.2 $\mathrm{L}$ at a temperature $27^{\circ} \mathrm{C}$ and a pressure of $760 \mathrm{~nm}$ of mercury in kilogram is [Molecular weight of oxygen $=32$ ]
139096
The tyre of a motor car contains air at $15^{\circ} \mathrm{C}$. If the temperature increases to $35^{\circ} \mathrm{C}$, the approximate percentage increase in pressure is (ignore to expansion of tyre)
138991
An ideal gas equation can be written as $p=\frac{\rho R T}{M_{0}}$ where, $\rho$ and $M_{0}$ are respectively,
1 mass density, mass of the gas
2 number density, molar mass
3 mass density, molar mass
4 number density, mass of the gas
Explanation:
C According to an ideal gas equation- $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{PV}=\frac{\mathrm{m}}{\mathrm{M}_{\mathrm{o}}} \mathrm{RT} \quad\left(\because \mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}_{\mathrm{o}}}\right)$ $\mathrm{P}=\left(\frac{\mathrm{m}}{\mathrm{V}}\right) \frac{\mathrm{RT}}{\mathrm{M}_{\mathrm{o}}}$ $\mathrm{P}=\rho \frac{\mathrm{RT}}{\mathrm{M}_{\mathrm{o}}} \quad\left(\because \rho=\frac{\mathrm{m}}{\mathrm{V}}\right)$ Where $\quad \rho=$ mass density $\mathrm{M}_{\mathrm{o}}=$ Molar mass
NEET OCT-2020
Kinetic Theory of Gases
139028
Pressure of an ideal gas is increased by keeping temperature constant. What is the effect on kinetic energy of molecules?
1 Increase
2 Decrease
3 No change
4 Can't be determined
Explanation:
C Kinetic energy of an ideal gas - $\text { K.E. }=\frac{3}{2} \mathrm{NK}_{\mathrm{B}} \mathrm{T}$ Where, $\mathrm{N}=$ Number of molecules $\mathrm{K}_{\mathrm{B}}=\text { Boltzmann's constant }$ Since kinetic energy is independent of pressure, therefore on increasing the pressure by keeping the temperature constant the kinetic energy of ideal gas remains constant. That means there will be no change on the kinetic energy of molecules.
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Kinetic Theory of Gases
139095
The mass of oxygen gas occupying a volume of 11.2 $\mathrm{L}$ at a temperature $27^{\circ} \mathrm{C}$ and a pressure of $760 \mathrm{~nm}$ of mercury in kilogram is [Molecular weight of oxygen $=32$ ]
139096
The tyre of a motor car contains air at $15^{\circ} \mathrm{C}$. If the temperature increases to $35^{\circ} \mathrm{C}$, the approximate percentage increase in pressure is (ignore to expansion of tyre)
138991
An ideal gas equation can be written as $p=\frac{\rho R T}{M_{0}}$ where, $\rho$ and $M_{0}$ are respectively,
1 mass density, mass of the gas
2 number density, molar mass
3 mass density, molar mass
4 number density, mass of the gas
Explanation:
C According to an ideal gas equation- $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{PV}=\frac{\mathrm{m}}{\mathrm{M}_{\mathrm{o}}} \mathrm{RT} \quad\left(\because \mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}_{\mathrm{o}}}\right)$ $\mathrm{P}=\left(\frac{\mathrm{m}}{\mathrm{V}}\right) \frac{\mathrm{RT}}{\mathrm{M}_{\mathrm{o}}}$ $\mathrm{P}=\rho \frac{\mathrm{RT}}{\mathrm{M}_{\mathrm{o}}} \quad\left(\because \rho=\frac{\mathrm{m}}{\mathrm{V}}\right)$ Where $\quad \rho=$ mass density $\mathrm{M}_{\mathrm{o}}=$ Molar mass
NEET OCT-2020
Kinetic Theory of Gases
139028
Pressure of an ideal gas is increased by keeping temperature constant. What is the effect on kinetic energy of molecules?
1 Increase
2 Decrease
3 No change
4 Can't be determined
Explanation:
C Kinetic energy of an ideal gas - $\text { K.E. }=\frac{3}{2} \mathrm{NK}_{\mathrm{B}} \mathrm{T}$ Where, $\mathrm{N}=$ Number of molecules $\mathrm{K}_{\mathrm{B}}=\text { Boltzmann's constant }$ Since kinetic energy is independent of pressure, therefore on increasing the pressure by keeping the temperature constant the kinetic energy of ideal gas remains constant. That means there will be no change on the kinetic energy of molecules.
