139066
The graph of pressure $P$ and $\left(\frac{1}{\text { Volume(V) }}\right)$ of 1 mole of an ideal gas at constant temperature is
1
2
3
4
Explanation:
B Graph between $\mathrm{P}$ and $(1 / \mathrm{V})$ at constant temperature for an ideal gas - According to Boyles law - $\mathrm{PV}=\text { constant }$ Then $\quad \mathrm{P} \propto \frac{1}{\mathrm{~V}}$ When pressure increases the volume is decreases so $\left(\frac{1}{\mathrm{~V}}\right)$ is also increases. Hence, graph is straight line passing through the origin at constant temperature.
UP CPMT-2002
Kinetic Theory of Gases
139067
The molecular weight of a gas is 44. The volume occupied by $2.2 \mathrm{~g}$ of this gas at $0^{\circ} \mathrm{C}$ and 2 atmospheric pressure will be
1 $2.8 \mathrm{~L}$
2 $0.56 \mathrm{~L}$
3 $5.6 \mathrm{~L}$
4 $44.8 \mathrm{~L}$
Explanation:
B Molecular weight of gas (M) $=44 \mathrm{~g} / \mathrm{mol}$ Mass of gas $(\mathrm{m})=2.2 \mathrm{~g}$ $\mathrm{T}=0^{\circ} \mathrm{C}=273 \mathrm{~K}$ $\mathrm{P}=2 \mathrm{~atm}=2 \times 1.01 \times 10^{5}=2.02 \times 10^{5} \mathrm{~Pa}$ $\mathrm{R}=8.314 \mathrm{~J} / \mathrm{mol}-\mathrm{K}$ Number of moles $(n)=\frac{m}{M}=\frac{2.2}{44}=0.05 \mathrm{~mol}$ From Ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{V}=\frac{\mathrm{nRT}}{\mathrm{P}}$ $=\frac{0.05 \times 8.314 \times 273}{2.02 \times 10^{5}}$ $=56.18 \times 10^{-5} \mathrm{~m}^{3}$ $=56.18 \times 10^{-2} \text { litre } \approx 0.56 \text { litre }$
UP CPMT-2001
Kinetic Theory of Gases
139068
An ideal gas follows a process described by the equation $P V^{2}=C$ from the initial $\left(P_{1}, V_{1}, T_{1}\right)$ to final $\left(P_{2}, V_{2}, T_{2}\right)$ thermodynamics states, where $C$ is a constant. Then:
1 If $\mathrm{P}_{1}>\mathrm{P}_{2}$ then $\mathrm{T}_{1} \lt \mathrm{T}_{2}$
2 If $V_{2}>V_{1}$ then $T_{2}>T_{1}$
3 If $\mathrm{V}_{2}>\mathrm{V}_{1}$ then $\mathrm{T}_{2} \lt \mathrm{T}_{1}$
4 If $\mathrm{P}_{1}>\mathrm{P}_{2}$ then $\mathrm{T}_{1}>\mathrm{T}_{2}$
Explanation:
C Given, $\mathrm{PV}^{2}=\mathrm{C} \quad$....(i) According to an ideal gas equation - $\mathrm{PV}=\mathrm{nRT}$ From equation (i) and (ii), we get - $\left(\frac{\mathrm{nRT}}{\mathrm{V}}\right) \mathrm{V}^{2}=\mathrm{C}$ $\mathrm{TV}=\mathrm{C}$ $\mathrm{T}_{1} \mathrm{~V}_{1}=\mathrm{T}_{2} \mathrm{~V}_{2}$ If temperature increase then volume decreases. Hence, $V_{2}>V_{1}$ then $T_{2} \lt T_{1}$.
Kinetic Theory of Gases
139069
The volume occupied by the molecules contained in $4.5 \mathrm{~kg}$ water at STP, if the intermolecular forces vanish away is
1 $5.6 \times 10^{-3} \mathrm{~m}^{3}$
2 $5.6 \mathrm{~m}^{3}$
3 $5.6 \times 10^{6} \mathrm{~m}^{3}$
4 $5.6 \times 10^{3} \mathrm{~m}^{3}$
Explanation:
B Given, weight of water $=4.5 \mathrm{~kg}$, molecular mass of water vapour $=18 \mathrm{gm}=18 \times 10^{-3} \mathrm{~kg}$ We know that, volume occupied by 1 mole of water vapour at $\mathrm{STP}=22.4$ litres $=22.4 \times 10^{-3} \mathrm{~m}^{3}$ $\therefore$ Volume occupied by $4.5 \mathrm{~kg}$ of water molecule $=$ Number of moles of water $\times(22.4$ litre $)$ $\mathrm{V}=\frac{\text { mass }}{\text { molar mass }} \times\left(22.4 \times 10^{-3} \mathrm{~m}^{3}\right)$ $\mathrm{V}=\frac{4.5}{18 \times 10^{-3}} \times 22.4 \times 10^{-3}$ $\mathrm{~V}=5.6 \mathrm{~m}^{3}$
139066
The graph of pressure $P$ and $\left(\frac{1}{\text { Volume(V) }}\right)$ of 1 mole of an ideal gas at constant temperature is
1
2
3
4
Explanation:
B Graph between $\mathrm{P}$ and $(1 / \mathrm{V})$ at constant temperature for an ideal gas - According to Boyles law - $\mathrm{PV}=\text { constant }$ Then $\quad \mathrm{P} \propto \frac{1}{\mathrm{~V}}$ When pressure increases the volume is decreases so $\left(\frac{1}{\mathrm{~V}}\right)$ is also increases. Hence, graph is straight line passing through the origin at constant temperature.
UP CPMT-2002
Kinetic Theory of Gases
139067
The molecular weight of a gas is 44. The volume occupied by $2.2 \mathrm{~g}$ of this gas at $0^{\circ} \mathrm{C}$ and 2 atmospheric pressure will be
1 $2.8 \mathrm{~L}$
2 $0.56 \mathrm{~L}$
3 $5.6 \mathrm{~L}$
4 $44.8 \mathrm{~L}$
Explanation:
B Molecular weight of gas (M) $=44 \mathrm{~g} / \mathrm{mol}$ Mass of gas $(\mathrm{m})=2.2 \mathrm{~g}$ $\mathrm{T}=0^{\circ} \mathrm{C}=273 \mathrm{~K}$ $\mathrm{P}=2 \mathrm{~atm}=2 \times 1.01 \times 10^{5}=2.02 \times 10^{5} \mathrm{~Pa}$ $\mathrm{R}=8.314 \mathrm{~J} / \mathrm{mol}-\mathrm{K}$ Number of moles $(n)=\frac{m}{M}=\frac{2.2}{44}=0.05 \mathrm{~mol}$ From Ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{V}=\frac{\mathrm{nRT}}{\mathrm{P}}$ $=\frac{0.05 \times 8.314 \times 273}{2.02 \times 10^{5}}$ $=56.18 \times 10^{-5} \mathrm{~m}^{3}$ $=56.18 \times 10^{-2} \text { litre } \approx 0.56 \text { litre }$
UP CPMT-2001
Kinetic Theory of Gases
139068
An ideal gas follows a process described by the equation $P V^{2}=C$ from the initial $\left(P_{1}, V_{1}, T_{1}\right)$ to final $\left(P_{2}, V_{2}, T_{2}\right)$ thermodynamics states, where $C$ is a constant. Then:
1 If $\mathrm{P}_{1}>\mathrm{P}_{2}$ then $\mathrm{T}_{1} \lt \mathrm{T}_{2}$
2 If $V_{2}>V_{1}$ then $T_{2}>T_{1}$
3 If $\mathrm{V}_{2}>\mathrm{V}_{1}$ then $\mathrm{T}_{2} \lt \mathrm{T}_{1}$
4 If $\mathrm{P}_{1}>\mathrm{P}_{2}$ then $\mathrm{T}_{1}>\mathrm{T}_{2}$
Explanation:
C Given, $\mathrm{PV}^{2}=\mathrm{C} \quad$....(i) According to an ideal gas equation - $\mathrm{PV}=\mathrm{nRT}$ From equation (i) and (ii), we get - $\left(\frac{\mathrm{nRT}}{\mathrm{V}}\right) \mathrm{V}^{2}=\mathrm{C}$ $\mathrm{TV}=\mathrm{C}$ $\mathrm{T}_{1} \mathrm{~V}_{1}=\mathrm{T}_{2} \mathrm{~V}_{2}$ If temperature increase then volume decreases. Hence, $V_{2}>V_{1}$ then $T_{2} \lt T_{1}$.
Kinetic Theory of Gases
139069
The volume occupied by the molecules contained in $4.5 \mathrm{~kg}$ water at STP, if the intermolecular forces vanish away is
1 $5.6 \times 10^{-3} \mathrm{~m}^{3}$
2 $5.6 \mathrm{~m}^{3}$
3 $5.6 \times 10^{6} \mathrm{~m}^{3}$
4 $5.6 \times 10^{3} \mathrm{~m}^{3}$
Explanation:
B Given, weight of water $=4.5 \mathrm{~kg}$, molecular mass of water vapour $=18 \mathrm{gm}=18 \times 10^{-3} \mathrm{~kg}$ We know that, volume occupied by 1 mole of water vapour at $\mathrm{STP}=22.4$ litres $=22.4 \times 10^{-3} \mathrm{~m}^{3}$ $\therefore$ Volume occupied by $4.5 \mathrm{~kg}$ of water molecule $=$ Number of moles of water $\times(22.4$ litre $)$ $\mathrm{V}=\frac{\text { mass }}{\text { molar mass }} \times\left(22.4 \times 10^{-3} \mathrm{~m}^{3}\right)$ $\mathrm{V}=\frac{4.5}{18 \times 10^{-3}} \times 22.4 \times 10^{-3}$ $\mathrm{~V}=5.6 \mathrm{~m}^{3}$
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Kinetic Theory of Gases
139066
The graph of pressure $P$ and $\left(\frac{1}{\text { Volume(V) }}\right)$ of 1 mole of an ideal gas at constant temperature is
1
2
3
4
Explanation:
B Graph between $\mathrm{P}$ and $(1 / \mathrm{V})$ at constant temperature for an ideal gas - According to Boyles law - $\mathrm{PV}=\text { constant }$ Then $\quad \mathrm{P} \propto \frac{1}{\mathrm{~V}}$ When pressure increases the volume is decreases so $\left(\frac{1}{\mathrm{~V}}\right)$ is also increases. Hence, graph is straight line passing through the origin at constant temperature.
UP CPMT-2002
Kinetic Theory of Gases
139067
The molecular weight of a gas is 44. The volume occupied by $2.2 \mathrm{~g}$ of this gas at $0^{\circ} \mathrm{C}$ and 2 atmospheric pressure will be
1 $2.8 \mathrm{~L}$
2 $0.56 \mathrm{~L}$
3 $5.6 \mathrm{~L}$
4 $44.8 \mathrm{~L}$
Explanation:
B Molecular weight of gas (M) $=44 \mathrm{~g} / \mathrm{mol}$ Mass of gas $(\mathrm{m})=2.2 \mathrm{~g}$ $\mathrm{T}=0^{\circ} \mathrm{C}=273 \mathrm{~K}$ $\mathrm{P}=2 \mathrm{~atm}=2 \times 1.01 \times 10^{5}=2.02 \times 10^{5} \mathrm{~Pa}$ $\mathrm{R}=8.314 \mathrm{~J} / \mathrm{mol}-\mathrm{K}$ Number of moles $(n)=\frac{m}{M}=\frac{2.2}{44}=0.05 \mathrm{~mol}$ From Ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{V}=\frac{\mathrm{nRT}}{\mathrm{P}}$ $=\frac{0.05 \times 8.314 \times 273}{2.02 \times 10^{5}}$ $=56.18 \times 10^{-5} \mathrm{~m}^{3}$ $=56.18 \times 10^{-2} \text { litre } \approx 0.56 \text { litre }$
UP CPMT-2001
Kinetic Theory of Gases
139068
An ideal gas follows a process described by the equation $P V^{2}=C$ from the initial $\left(P_{1}, V_{1}, T_{1}\right)$ to final $\left(P_{2}, V_{2}, T_{2}\right)$ thermodynamics states, where $C$ is a constant. Then:
1 If $\mathrm{P}_{1}>\mathrm{P}_{2}$ then $\mathrm{T}_{1} \lt \mathrm{T}_{2}$
2 If $V_{2}>V_{1}$ then $T_{2}>T_{1}$
3 If $\mathrm{V}_{2}>\mathrm{V}_{1}$ then $\mathrm{T}_{2} \lt \mathrm{T}_{1}$
4 If $\mathrm{P}_{1}>\mathrm{P}_{2}$ then $\mathrm{T}_{1}>\mathrm{T}_{2}$
Explanation:
C Given, $\mathrm{PV}^{2}=\mathrm{C} \quad$....(i) According to an ideal gas equation - $\mathrm{PV}=\mathrm{nRT}$ From equation (i) and (ii), we get - $\left(\frac{\mathrm{nRT}}{\mathrm{V}}\right) \mathrm{V}^{2}=\mathrm{C}$ $\mathrm{TV}=\mathrm{C}$ $\mathrm{T}_{1} \mathrm{~V}_{1}=\mathrm{T}_{2} \mathrm{~V}_{2}$ If temperature increase then volume decreases. Hence, $V_{2}>V_{1}$ then $T_{2} \lt T_{1}$.
Kinetic Theory of Gases
139069
The volume occupied by the molecules contained in $4.5 \mathrm{~kg}$ water at STP, if the intermolecular forces vanish away is
1 $5.6 \times 10^{-3} \mathrm{~m}^{3}$
2 $5.6 \mathrm{~m}^{3}$
3 $5.6 \times 10^{6} \mathrm{~m}^{3}$
4 $5.6 \times 10^{3} \mathrm{~m}^{3}$
Explanation:
B Given, weight of water $=4.5 \mathrm{~kg}$, molecular mass of water vapour $=18 \mathrm{gm}=18 \times 10^{-3} \mathrm{~kg}$ We know that, volume occupied by 1 mole of water vapour at $\mathrm{STP}=22.4$ litres $=22.4 \times 10^{-3} \mathrm{~m}^{3}$ $\therefore$ Volume occupied by $4.5 \mathrm{~kg}$ of water molecule $=$ Number of moles of water $\times(22.4$ litre $)$ $\mathrm{V}=\frac{\text { mass }}{\text { molar mass }} \times\left(22.4 \times 10^{-3} \mathrm{~m}^{3}\right)$ $\mathrm{V}=\frac{4.5}{18 \times 10^{-3}} \times 22.4 \times 10^{-3}$ $\mathrm{~V}=5.6 \mathrm{~m}^{3}$
139066
The graph of pressure $P$ and $\left(\frac{1}{\text { Volume(V) }}\right)$ of 1 mole of an ideal gas at constant temperature is
1
2
3
4
Explanation:
B Graph between $\mathrm{P}$ and $(1 / \mathrm{V})$ at constant temperature for an ideal gas - According to Boyles law - $\mathrm{PV}=\text { constant }$ Then $\quad \mathrm{P} \propto \frac{1}{\mathrm{~V}}$ When pressure increases the volume is decreases so $\left(\frac{1}{\mathrm{~V}}\right)$ is also increases. Hence, graph is straight line passing through the origin at constant temperature.
UP CPMT-2002
Kinetic Theory of Gases
139067
The molecular weight of a gas is 44. The volume occupied by $2.2 \mathrm{~g}$ of this gas at $0^{\circ} \mathrm{C}$ and 2 atmospheric pressure will be
1 $2.8 \mathrm{~L}$
2 $0.56 \mathrm{~L}$
3 $5.6 \mathrm{~L}$
4 $44.8 \mathrm{~L}$
Explanation:
B Molecular weight of gas (M) $=44 \mathrm{~g} / \mathrm{mol}$ Mass of gas $(\mathrm{m})=2.2 \mathrm{~g}$ $\mathrm{T}=0^{\circ} \mathrm{C}=273 \mathrm{~K}$ $\mathrm{P}=2 \mathrm{~atm}=2 \times 1.01 \times 10^{5}=2.02 \times 10^{5} \mathrm{~Pa}$ $\mathrm{R}=8.314 \mathrm{~J} / \mathrm{mol}-\mathrm{K}$ Number of moles $(n)=\frac{m}{M}=\frac{2.2}{44}=0.05 \mathrm{~mol}$ From Ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{V}=\frac{\mathrm{nRT}}{\mathrm{P}}$ $=\frac{0.05 \times 8.314 \times 273}{2.02 \times 10^{5}}$ $=56.18 \times 10^{-5} \mathrm{~m}^{3}$ $=56.18 \times 10^{-2} \text { litre } \approx 0.56 \text { litre }$
UP CPMT-2001
Kinetic Theory of Gases
139068
An ideal gas follows a process described by the equation $P V^{2}=C$ from the initial $\left(P_{1}, V_{1}, T_{1}\right)$ to final $\left(P_{2}, V_{2}, T_{2}\right)$ thermodynamics states, where $C$ is a constant. Then:
1 If $\mathrm{P}_{1}>\mathrm{P}_{2}$ then $\mathrm{T}_{1} \lt \mathrm{T}_{2}$
2 If $V_{2}>V_{1}$ then $T_{2}>T_{1}$
3 If $\mathrm{V}_{2}>\mathrm{V}_{1}$ then $\mathrm{T}_{2} \lt \mathrm{T}_{1}$
4 If $\mathrm{P}_{1}>\mathrm{P}_{2}$ then $\mathrm{T}_{1}>\mathrm{T}_{2}$
Explanation:
C Given, $\mathrm{PV}^{2}=\mathrm{C} \quad$....(i) According to an ideal gas equation - $\mathrm{PV}=\mathrm{nRT}$ From equation (i) and (ii), we get - $\left(\frac{\mathrm{nRT}}{\mathrm{V}}\right) \mathrm{V}^{2}=\mathrm{C}$ $\mathrm{TV}=\mathrm{C}$ $\mathrm{T}_{1} \mathrm{~V}_{1}=\mathrm{T}_{2} \mathrm{~V}_{2}$ If temperature increase then volume decreases. Hence, $V_{2}>V_{1}$ then $T_{2} \lt T_{1}$.
Kinetic Theory of Gases
139069
The volume occupied by the molecules contained in $4.5 \mathrm{~kg}$ water at STP, if the intermolecular forces vanish away is
1 $5.6 \times 10^{-3} \mathrm{~m}^{3}$
2 $5.6 \mathrm{~m}^{3}$
3 $5.6 \times 10^{6} \mathrm{~m}^{3}$
4 $5.6 \times 10^{3} \mathrm{~m}^{3}$
Explanation:
B Given, weight of water $=4.5 \mathrm{~kg}$, molecular mass of water vapour $=18 \mathrm{gm}=18 \times 10^{-3} \mathrm{~kg}$ We know that, volume occupied by 1 mole of water vapour at $\mathrm{STP}=22.4$ litres $=22.4 \times 10^{-3} \mathrm{~m}^{3}$ $\therefore$ Volume occupied by $4.5 \mathrm{~kg}$ of water molecule $=$ Number of moles of water $\times(22.4$ litre $)$ $\mathrm{V}=\frac{\text { mass }}{\text { molar mass }} \times\left(22.4 \times 10^{-3} \mathrm{~m}^{3}\right)$ $\mathrm{V}=\frac{4.5}{18 \times 10^{-3}} \times 22.4 \times 10^{-3}$ $\mathrm{~V}=5.6 \mathrm{~m}^{3}$
