139006
In a given process on an ideal gas, $d W=0$ and $d Q \lt 0$. Then for the gas
1 the temperature will decrease
2 the volume will increase
3 the pressure will remain constant
4 the temperature will increase
Explanation:
A Given that, $\mathrm{dW}=0$ and $\mathrm{dQ} \lt 0$ The first law of thermodynamics $\mathrm{dQ}=\mathrm{dU}+\mathrm{dW}$ $\mathrm{dQ}=\mathrm{dU}+0$ $\mathrm{dQ}=\mathrm{dU}$ $\mathrm{dQ} \lt 0 \text { then } \mathrm{dU} \lt 0$ We know that, $\mathrm{dU} =\mathrm{nC}_{\mathrm{V}} \mathrm{dT}$ $=\mathrm{nC}_{\mathrm{V}}\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right) \lt 0$ $\mathrm{nC}_{\mathrm{V}} =\text { always positive }$ Then, $\quad \mathrm{T}_{2} \lt \mathrm{T}_{1}$ So, the temperature will decrease.
HP CET-2018
Kinetic Theory of Gases
139007
A gaseous mixture has 2 moles of oxygen and 4 moles of Argon at a temperature T. Neglecting all vibrational modes of the molecules, the total internal energy of the system is (R-Universal gas constant)
1 $4 \mathrm{RT}$
2 $15 \mathrm{RT}$
3 9 RT
4 $11 \mathrm{RT}$
Explanation:
D Given that, $\mathrm{O}_{2} \rightarrow 2$ mole $=\mathrm{n}_{1}$ (diatomic gas) $\mathrm{Ar} \rightarrow 4$ mole $=\mathrm{n}_{2}$ (Monoatomic gas) Then degree of freedom for diatomic, $f=5$ and degree of freedom for monoatomic, $f=3$ Then, total internal Energy $\mathrm{U}=\mathrm{n}_1 \mathrm{C}_{\mathrm{V}} \mathrm{dT}+\mathrm{n}_2 \mathrm{C}_{\mathrm{V}} \mathrm{dT}$ $\mathrm{U}=\mathrm{n}_1 \frac{\mathrm{f}}{2} \mathrm{RT}+\mathrm{n}_2 \frac{\mathrm{f}}{2} \mathrm{RT}$ $\mathrm{U}=2 \times \frac{5}{2} \mathrm{RT}+4 \times \frac{3}{2} \mathrm{RT}$ $\mathrm{U}=5 \mathrm{RT}+6 \mathrm{RT}=11 \mathrm{RT}$ $\mathrm{U}=11 \mathrm{RT}$
Shift-II
Kinetic Theory of Gases
139008
Which of the graphs shown in the following figures correctly represents the variation of $\beta=-\left(\frac{d V}{d P}\right) \frac{1}{V}$ with $P$ for an ideal gas at constant temperature?
1
2
3
4
Explanation:
A According to Boyle's law $\mathrm{PV}=$ Constant Partial differentiation $\mathrm{PdV}+\mathrm{VdP}=0$ $P=-V\left(\frac{d P}{d V}\right)$ $\Rightarrow \quad \frac{1}{\mathrm{P}}=-\frac{1}{\mathrm{~V}}\left(\frac{\mathrm{dV}}{\mathrm{dP}}\right)=\beta$ So, $\quad \beta \propto \frac{1}{\mathrm{P}}$ So, When pressure increase, then $\beta$ decrease.
2014]
Kinetic Theory of Gases
139009
If a kilo mole of methane gas weights $16 \mathrm{~kg}$ the density of methane at $20^{\circ} \mathrm{C}$ and $5 \mathrm{~atm}$ pressure is. (Use R $=8.314 \mathrm{~J} / \mathrm{mol}-\mathrm{K})$
139006
In a given process on an ideal gas, $d W=0$ and $d Q \lt 0$. Then for the gas
1 the temperature will decrease
2 the volume will increase
3 the pressure will remain constant
4 the temperature will increase
Explanation:
A Given that, $\mathrm{dW}=0$ and $\mathrm{dQ} \lt 0$ The first law of thermodynamics $\mathrm{dQ}=\mathrm{dU}+\mathrm{dW}$ $\mathrm{dQ}=\mathrm{dU}+0$ $\mathrm{dQ}=\mathrm{dU}$ $\mathrm{dQ} \lt 0 \text { then } \mathrm{dU} \lt 0$ We know that, $\mathrm{dU} =\mathrm{nC}_{\mathrm{V}} \mathrm{dT}$ $=\mathrm{nC}_{\mathrm{V}}\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right) \lt 0$ $\mathrm{nC}_{\mathrm{V}} =\text { always positive }$ Then, $\quad \mathrm{T}_{2} \lt \mathrm{T}_{1}$ So, the temperature will decrease.
HP CET-2018
Kinetic Theory of Gases
139007
A gaseous mixture has 2 moles of oxygen and 4 moles of Argon at a temperature T. Neglecting all vibrational modes of the molecules, the total internal energy of the system is (R-Universal gas constant)
1 $4 \mathrm{RT}$
2 $15 \mathrm{RT}$
3 9 RT
4 $11 \mathrm{RT}$
Explanation:
D Given that, $\mathrm{O}_{2} \rightarrow 2$ mole $=\mathrm{n}_{1}$ (diatomic gas) $\mathrm{Ar} \rightarrow 4$ mole $=\mathrm{n}_{2}$ (Monoatomic gas) Then degree of freedom for diatomic, $f=5$ and degree of freedom for monoatomic, $f=3$ Then, total internal Energy $\mathrm{U}=\mathrm{n}_1 \mathrm{C}_{\mathrm{V}} \mathrm{dT}+\mathrm{n}_2 \mathrm{C}_{\mathrm{V}} \mathrm{dT}$ $\mathrm{U}=\mathrm{n}_1 \frac{\mathrm{f}}{2} \mathrm{RT}+\mathrm{n}_2 \frac{\mathrm{f}}{2} \mathrm{RT}$ $\mathrm{U}=2 \times \frac{5}{2} \mathrm{RT}+4 \times \frac{3}{2} \mathrm{RT}$ $\mathrm{U}=5 \mathrm{RT}+6 \mathrm{RT}=11 \mathrm{RT}$ $\mathrm{U}=11 \mathrm{RT}$
Shift-II
Kinetic Theory of Gases
139008
Which of the graphs shown in the following figures correctly represents the variation of $\beta=-\left(\frac{d V}{d P}\right) \frac{1}{V}$ with $P$ for an ideal gas at constant temperature?
1
2
3
4
Explanation:
A According to Boyle's law $\mathrm{PV}=$ Constant Partial differentiation $\mathrm{PdV}+\mathrm{VdP}=0$ $P=-V\left(\frac{d P}{d V}\right)$ $\Rightarrow \quad \frac{1}{\mathrm{P}}=-\frac{1}{\mathrm{~V}}\left(\frac{\mathrm{dV}}{\mathrm{dP}}\right)=\beta$ So, $\quad \beta \propto \frac{1}{\mathrm{P}}$ So, When pressure increase, then $\beta$ decrease.
2014]
Kinetic Theory of Gases
139009
If a kilo mole of methane gas weights $16 \mathrm{~kg}$ the density of methane at $20^{\circ} \mathrm{C}$ and $5 \mathrm{~atm}$ pressure is. (Use R $=8.314 \mathrm{~J} / \mathrm{mol}-\mathrm{K})$
139006
In a given process on an ideal gas, $d W=0$ and $d Q \lt 0$. Then for the gas
1 the temperature will decrease
2 the volume will increase
3 the pressure will remain constant
4 the temperature will increase
Explanation:
A Given that, $\mathrm{dW}=0$ and $\mathrm{dQ} \lt 0$ The first law of thermodynamics $\mathrm{dQ}=\mathrm{dU}+\mathrm{dW}$ $\mathrm{dQ}=\mathrm{dU}+0$ $\mathrm{dQ}=\mathrm{dU}$ $\mathrm{dQ} \lt 0 \text { then } \mathrm{dU} \lt 0$ We know that, $\mathrm{dU} =\mathrm{nC}_{\mathrm{V}} \mathrm{dT}$ $=\mathrm{nC}_{\mathrm{V}}\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right) \lt 0$ $\mathrm{nC}_{\mathrm{V}} =\text { always positive }$ Then, $\quad \mathrm{T}_{2} \lt \mathrm{T}_{1}$ So, the temperature will decrease.
HP CET-2018
Kinetic Theory of Gases
139007
A gaseous mixture has 2 moles of oxygen and 4 moles of Argon at a temperature T. Neglecting all vibrational modes of the molecules, the total internal energy of the system is (R-Universal gas constant)
1 $4 \mathrm{RT}$
2 $15 \mathrm{RT}$
3 9 RT
4 $11 \mathrm{RT}$
Explanation:
D Given that, $\mathrm{O}_{2} \rightarrow 2$ mole $=\mathrm{n}_{1}$ (diatomic gas) $\mathrm{Ar} \rightarrow 4$ mole $=\mathrm{n}_{2}$ (Monoatomic gas) Then degree of freedom for diatomic, $f=5$ and degree of freedom for monoatomic, $f=3$ Then, total internal Energy $\mathrm{U}=\mathrm{n}_1 \mathrm{C}_{\mathrm{V}} \mathrm{dT}+\mathrm{n}_2 \mathrm{C}_{\mathrm{V}} \mathrm{dT}$ $\mathrm{U}=\mathrm{n}_1 \frac{\mathrm{f}}{2} \mathrm{RT}+\mathrm{n}_2 \frac{\mathrm{f}}{2} \mathrm{RT}$ $\mathrm{U}=2 \times \frac{5}{2} \mathrm{RT}+4 \times \frac{3}{2} \mathrm{RT}$ $\mathrm{U}=5 \mathrm{RT}+6 \mathrm{RT}=11 \mathrm{RT}$ $\mathrm{U}=11 \mathrm{RT}$
Shift-II
Kinetic Theory of Gases
139008
Which of the graphs shown in the following figures correctly represents the variation of $\beta=-\left(\frac{d V}{d P}\right) \frac{1}{V}$ with $P$ for an ideal gas at constant temperature?
1
2
3
4
Explanation:
A According to Boyle's law $\mathrm{PV}=$ Constant Partial differentiation $\mathrm{PdV}+\mathrm{VdP}=0$ $P=-V\left(\frac{d P}{d V}\right)$ $\Rightarrow \quad \frac{1}{\mathrm{P}}=-\frac{1}{\mathrm{~V}}\left(\frac{\mathrm{dV}}{\mathrm{dP}}\right)=\beta$ So, $\quad \beta \propto \frac{1}{\mathrm{P}}$ So, When pressure increase, then $\beta$ decrease.
2014]
Kinetic Theory of Gases
139009
If a kilo mole of methane gas weights $16 \mathrm{~kg}$ the density of methane at $20^{\circ} \mathrm{C}$ and $5 \mathrm{~atm}$ pressure is. (Use R $=8.314 \mathrm{~J} / \mathrm{mol}-\mathrm{K})$
139006
In a given process on an ideal gas, $d W=0$ and $d Q \lt 0$. Then for the gas
1 the temperature will decrease
2 the volume will increase
3 the pressure will remain constant
4 the temperature will increase
Explanation:
A Given that, $\mathrm{dW}=0$ and $\mathrm{dQ} \lt 0$ The first law of thermodynamics $\mathrm{dQ}=\mathrm{dU}+\mathrm{dW}$ $\mathrm{dQ}=\mathrm{dU}+0$ $\mathrm{dQ}=\mathrm{dU}$ $\mathrm{dQ} \lt 0 \text { then } \mathrm{dU} \lt 0$ We know that, $\mathrm{dU} =\mathrm{nC}_{\mathrm{V}} \mathrm{dT}$ $=\mathrm{nC}_{\mathrm{V}}\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right) \lt 0$ $\mathrm{nC}_{\mathrm{V}} =\text { always positive }$ Then, $\quad \mathrm{T}_{2} \lt \mathrm{T}_{1}$ So, the temperature will decrease.
HP CET-2018
Kinetic Theory of Gases
139007
A gaseous mixture has 2 moles of oxygen and 4 moles of Argon at a temperature T. Neglecting all vibrational modes of the molecules, the total internal energy of the system is (R-Universal gas constant)
1 $4 \mathrm{RT}$
2 $15 \mathrm{RT}$
3 9 RT
4 $11 \mathrm{RT}$
Explanation:
D Given that, $\mathrm{O}_{2} \rightarrow 2$ mole $=\mathrm{n}_{1}$ (diatomic gas) $\mathrm{Ar} \rightarrow 4$ mole $=\mathrm{n}_{2}$ (Monoatomic gas) Then degree of freedom for diatomic, $f=5$ and degree of freedom for monoatomic, $f=3$ Then, total internal Energy $\mathrm{U}=\mathrm{n}_1 \mathrm{C}_{\mathrm{V}} \mathrm{dT}+\mathrm{n}_2 \mathrm{C}_{\mathrm{V}} \mathrm{dT}$ $\mathrm{U}=\mathrm{n}_1 \frac{\mathrm{f}}{2} \mathrm{RT}+\mathrm{n}_2 \frac{\mathrm{f}}{2} \mathrm{RT}$ $\mathrm{U}=2 \times \frac{5}{2} \mathrm{RT}+4 \times \frac{3}{2} \mathrm{RT}$ $\mathrm{U}=5 \mathrm{RT}+6 \mathrm{RT}=11 \mathrm{RT}$ $\mathrm{U}=11 \mathrm{RT}$
Shift-II
Kinetic Theory of Gases
139008
Which of the graphs shown in the following figures correctly represents the variation of $\beta=-\left(\frac{d V}{d P}\right) \frac{1}{V}$ with $P$ for an ideal gas at constant temperature?
1
2
3
4
Explanation:
A According to Boyle's law $\mathrm{PV}=$ Constant Partial differentiation $\mathrm{PdV}+\mathrm{VdP}=0$ $P=-V\left(\frac{d P}{d V}\right)$ $\Rightarrow \quad \frac{1}{\mathrm{P}}=-\frac{1}{\mathrm{~V}}\left(\frac{\mathrm{dV}}{\mathrm{dP}}\right)=\beta$ So, $\quad \beta \propto \frac{1}{\mathrm{P}}$ So, When pressure increase, then $\beta$ decrease.
2014]
Kinetic Theory of Gases
139009
If a kilo mole of methane gas weights $16 \mathrm{~kg}$ the density of methane at $20^{\circ} \mathrm{C}$ and $5 \mathrm{~atm}$ pressure is. (Use R $=8.314 \mathrm{~J} / \mathrm{mol}-\mathrm{K})$
