3 Triple temperature is $2 \mathrm{a} / \mathrm{Rb}$
4 Inversion temperature is $\mathrm{a} / 2 \mathrm{Rb}$
Explanation:
A Boyle's temperature is defined as the point in the temperature range in which a real gas starts to behave like an ideal gas at a pressure range. The temperature at which the second coefficient in the expression becomes zero which is called as Boyle temperature, $\therefore \quad \mathrm{T}_{\mathrm{B}}=\frac{\mathrm{a}}{\mathrm{Rb}}$
CG PET- 2009
Kinetic Theory of Gases
139043
The variation of $P V$ with $V$ of a fixed mass of an ideal gas at constant temperature is graphically represented by:
1
2
3
4
Explanation:
D According to Ideal gas equation- $\mathrm{PV}=\mathrm{nRT}$ Since, $\mathrm{PV}=$ constant Hence, for any value of V, PV remains constant The correct variation of $\mathrm{PV}$ with respect to $\mathrm{V}$ is -
BITSAT-2009
Kinetic Theory of Gases
139044
The universal gas law $\left(\frac{P V}{T}=\right.$ constant $)$ is applicable to
1 isothermal changes only
2 adiabatic changes only
3 both isothermal adiabatic changes
4 neither isothermal adiabatic changes
Explanation:
C According to ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$ $\frac{\mathrm{PV}}{\mathrm{T}}=\mathrm{nR}$ $\frac{\mathrm{PV}}{\mathrm{T}}=\text { constant }$ It is applicable to both isothermal and adiabatic changes.
COMEDK 2014
Kinetic Theory of Gases
139050
Which one of the following quantities is zero on an average for the molecules of an ideal gas in equilibrium?
1 Kinetic energy
2 Density
3 Speed
4 Momentum
Explanation:
C We know that, for an ideal gas $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}$ And $\quad C_{P}=\frac{d Q}{m \Delta T}$ $\mathrm{C}_{\mathrm{P}}=\frac{\mathrm{dU}+\mathrm{dW}}{\mathrm{m} \Delta \mathrm{T}} \quad(\because \mathrm{dQ}=\mathrm{dU}+\mathrm{dW})$ $\mathrm{C}_{\mathrm{P}}=\frac{\mathrm{dU}}{\mathrm{m} \Delta \mathrm{T}}+\frac{\mathrm{dW}}{\mathrm{m} \Delta \mathrm{T}} \ldots \ldots \text { (i) }$ Since, $C_{V}=\frac{d Q}{m \Delta T}$ $C_{V}=\frac{d U+d W}{m \Delta T}$ $C_{v}=\frac{d W}{m \Delta T}$ From equation (i) and equation (ii), we get- $\mathrm{C}_{\mathrm{P}}=\frac{\mathrm{dU}}{\mathrm{m} \Delta \mathrm{T}}+\mathrm{C}_{\mathrm{V}}$ So, we can say that $C_{P}$ is greater than $C_{V}$.
3 Triple temperature is $2 \mathrm{a} / \mathrm{Rb}$
4 Inversion temperature is $\mathrm{a} / 2 \mathrm{Rb}$
Explanation:
A Boyle's temperature is defined as the point in the temperature range in which a real gas starts to behave like an ideal gas at a pressure range. The temperature at which the second coefficient in the expression becomes zero which is called as Boyle temperature, $\therefore \quad \mathrm{T}_{\mathrm{B}}=\frac{\mathrm{a}}{\mathrm{Rb}}$
CG PET- 2009
Kinetic Theory of Gases
139043
The variation of $P V$ with $V$ of a fixed mass of an ideal gas at constant temperature is graphically represented by:
1
2
3
4
Explanation:
D According to Ideal gas equation- $\mathrm{PV}=\mathrm{nRT}$ Since, $\mathrm{PV}=$ constant Hence, for any value of V, PV remains constant The correct variation of $\mathrm{PV}$ with respect to $\mathrm{V}$ is -
BITSAT-2009
Kinetic Theory of Gases
139044
The universal gas law $\left(\frac{P V}{T}=\right.$ constant $)$ is applicable to
1 isothermal changes only
2 adiabatic changes only
3 both isothermal adiabatic changes
4 neither isothermal adiabatic changes
Explanation:
C According to ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$ $\frac{\mathrm{PV}}{\mathrm{T}}=\mathrm{nR}$ $\frac{\mathrm{PV}}{\mathrm{T}}=\text { constant }$ It is applicable to both isothermal and adiabatic changes.
COMEDK 2014
Kinetic Theory of Gases
139050
Which one of the following quantities is zero on an average for the molecules of an ideal gas in equilibrium?
1 Kinetic energy
2 Density
3 Speed
4 Momentum
Explanation:
C We know that, for an ideal gas $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}$ And $\quad C_{P}=\frac{d Q}{m \Delta T}$ $\mathrm{C}_{\mathrm{P}}=\frac{\mathrm{dU}+\mathrm{dW}}{\mathrm{m} \Delta \mathrm{T}} \quad(\because \mathrm{dQ}=\mathrm{dU}+\mathrm{dW})$ $\mathrm{C}_{\mathrm{P}}=\frac{\mathrm{dU}}{\mathrm{m} \Delta \mathrm{T}}+\frac{\mathrm{dW}}{\mathrm{m} \Delta \mathrm{T}} \ldots \ldots \text { (i) }$ Since, $C_{V}=\frac{d Q}{m \Delta T}$ $C_{V}=\frac{d U+d W}{m \Delta T}$ $C_{v}=\frac{d W}{m \Delta T}$ From equation (i) and equation (ii), we get- $\mathrm{C}_{\mathrm{P}}=\frac{\mathrm{dU}}{\mathrm{m} \Delta \mathrm{T}}+\mathrm{C}_{\mathrm{V}}$ So, we can say that $C_{P}$ is greater than $C_{V}$.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Kinetic Theory of Gases
139034
For a real gas (Vander Walls' gas)
1 Boyle temperature is $a / R b$ )
2 Critical temperature is $a / R b$
3 Triple temperature is $2 \mathrm{a} / \mathrm{Rb}$
4 Inversion temperature is $\mathrm{a} / 2 \mathrm{Rb}$
Explanation:
A Boyle's temperature is defined as the point in the temperature range in which a real gas starts to behave like an ideal gas at a pressure range. The temperature at which the second coefficient in the expression becomes zero which is called as Boyle temperature, $\therefore \quad \mathrm{T}_{\mathrm{B}}=\frac{\mathrm{a}}{\mathrm{Rb}}$
CG PET- 2009
Kinetic Theory of Gases
139043
The variation of $P V$ with $V$ of a fixed mass of an ideal gas at constant temperature is graphically represented by:
1
2
3
4
Explanation:
D According to Ideal gas equation- $\mathrm{PV}=\mathrm{nRT}$ Since, $\mathrm{PV}=$ constant Hence, for any value of V, PV remains constant The correct variation of $\mathrm{PV}$ with respect to $\mathrm{V}$ is -
BITSAT-2009
Kinetic Theory of Gases
139044
The universal gas law $\left(\frac{P V}{T}=\right.$ constant $)$ is applicable to
1 isothermal changes only
2 adiabatic changes only
3 both isothermal adiabatic changes
4 neither isothermal adiabatic changes
Explanation:
C According to ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$ $\frac{\mathrm{PV}}{\mathrm{T}}=\mathrm{nR}$ $\frac{\mathrm{PV}}{\mathrm{T}}=\text { constant }$ It is applicable to both isothermal and adiabatic changes.
COMEDK 2014
Kinetic Theory of Gases
139050
Which one of the following quantities is zero on an average for the molecules of an ideal gas in equilibrium?
1 Kinetic energy
2 Density
3 Speed
4 Momentum
Explanation:
C We know that, for an ideal gas $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}$ And $\quad C_{P}=\frac{d Q}{m \Delta T}$ $\mathrm{C}_{\mathrm{P}}=\frac{\mathrm{dU}+\mathrm{dW}}{\mathrm{m} \Delta \mathrm{T}} \quad(\because \mathrm{dQ}=\mathrm{dU}+\mathrm{dW})$ $\mathrm{C}_{\mathrm{P}}=\frac{\mathrm{dU}}{\mathrm{m} \Delta \mathrm{T}}+\frac{\mathrm{dW}}{\mathrm{m} \Delta \mathrm{T}} \ldots \ldots \text { (i) }$ Since, $C_{V}=\frac{d Q}{m \Delta T}$ $C_{V}=\frac{d U+d W}{m \Delta T}$ $C_{v}=\frac{d W}{m \Delta T}$ From equation (i) and equation (ii), we get- $\mathrm{C}_{\mathrm{P}}=\frac{\mathrm{dU}}{\mathrm{m} \Delta \mathrm{T}}+\mathrm{C}_{\mathrm{V}}$ So, we can say that $C_{P}$ is greater than $C_{V}$.
3 Triple temperature is $2 \mathrm{a} / \mathrm{Rb}$
4 Inversion temperature is $\mathrm{a} / 2 \mathrm{Rb}$
Explanation:
A Boyle's temperature is defined as the point in the temperature range in which a real gas starts to behave like an ideal gas at a pressure range. The temperature at which the second coefficient in the expression becomes zero which is called as Boyle temperature, $\therefore \quad \mathrm{T}_{\mathrm{B}}=\frac{\mathrm{a}}{\mathrm{Rb}}$
CG PET- 2009
Kinetic Theory of Gases
139043
The variation of $P V$ with $V$ of a fixed mass of an ideal gas at constant temperature is graphically represented by:
1
2
3
4
Explanation:
D According to Ideal gas equation- $\mathrm{PV}=\mathrm{nRT}$ Since, $\mathrm{PV}=$ constant Hence, for any value of V, PV remains constant The correct variation of $\mathrm{PV}$ with respect to $\mathrm{V}$ is -
BITSAT-2009
Kinetic Theory of Gases
139044
The universal gas law $\left(\frac{P V}{T}=\right.$ constant $)$ is applicable to
1 isothermal changes only
2 adiabatic changes only
3 both isothermal adiabatic changes
4 neither isothermal adiabatic changes
Explanation:
C According to ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$ $\frac{\mathrm{PV}}{\mathrm{T}}=\mathrm{nR}$ $\frac{\mathrm{PV}}{\mathrm{T}}=\text { constant }$ It is applicable to both isothermal and adiabatic changes.
COMEDK 2014
Kinetic Theory of Gases
139050
Which one of the following quantities is zero on an average for the molecules of an ideal gas in equilibrium?
1 Kinetic energy
2 Density
3 Speed
4 Momentum
Explanation:
C We know that, for an ideal gas $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}$ And $\quad C_{P}=\frac{d Q}{m \Delta T}$ $\mathrm{C}_{\mathrm{P}}=\frac{\mathrm{dU}+\mathrm{dW}}{\mathrm{m} \Delta \mathrm{T}} \quad(\because \mathrm{dQ}=\mathrm{dU}+\mathrm{dW})$ $\mathrm{C}_{\mathrm{P}}=\frac{\mathrm{dU}}{\mathrm{m} \Delta \mathrm{T}}+\frac{\mathrm{dW}}{\mathrm{m} \Delta \mathrm{T}} \ldots \ldots \text { (i) }$ Since, $C_{V}=\frac{d Q}{m \Delta T}$ $C_{V}=\frac{d U+d W}{m \Delta T}$ $C_{v}=\frac{d W}{m \Delta T}$ From equation (i) and equation (ii), we get- $\mathrm{C}_{\mathrm{P}}=\frac{\mathrm{dU}}{\mathrm{m} \Delta \mathrm{T}}+\mathrm{C}_{\mathrm{V}}$ So, we can say that $C_{P}$ is greater than $C_{V}$.
