138968
A balloon contains $1500 \mathrm{~m}^{3}$ of helium at $27^{\circ} \mathrm{C}$ and 4 atmospheric pressure. The volume of helium at $-3^{\circ} \mathrm{C}$ temperature and 2 atmospheric pressure will,
138969
If pressure of $\mathrm{CO}_{2}$ (real gas) in a container is given by $P=\frac{R T}{2 V-b}-\frac{a}{4 b^{2}}$, then mass of the gas in container is-
1 $11 \mathrm{~g}$
2 $22 \mathrm{~g}$
3 $33 \mathrm{~g}$
4 $44 \mathrm{~g}$
Explanation:
B Given, Pressure of $\mathrm{CO}_{2}$ (real gas) $\mathrm{P}=\frac{\mathrm{RT}}{2 \mathrm{~V}-\mathrm{b}}-\frac{\mathrm{a}}{4 \mathrm{~b}^{2}}$ We know that, The Vander waal's gas equation, $\left[\mathrm{P}+\frac{\mathrm{an}^{2}}{\mathrm{~V}^{2}}\right][\mathrm{V}-\mathrm{nb}]=\mathrm{nRT}$ $\Rightarrow \quad \mathrm{P}=\left[\frac{\mathrm{nRT}}{\mathrm{V}-\mathrm{nb}}\right]-\frac{\mathrm{n}^{2} \mathrm{a}}{\mathrm{V}^{2}}$ Comparing the equation of pressure of $\mathrm{CO}_{2}$ and Vander waal's gas equation, $\frac{\mathrm{RT}}{2 \mathrm{~V}-\mathrm{b}}-\frac{\mathrm{a}}{4 \mathrm{~b}^{2}}=\frac{\mathrm{nRT}}{\mathrm{V}-\mathrm{nb}}-\frac{\mathrm{n}^{2} \mathrm{a}}{\mathrm{V}^{2}}$ We get, $\mathrm{n}=\frac{1}{2}$ Again we know, The value of the mass $(\mathrm{m})$ $=$ molar mass $(M) \times$ Number of moles $(n)$ $\mathrm{m}=\mathrm{M} . \mathrm{n}$ Molar mass of $\mathrm{CO}_{2}=44 \mathrm{~g} / \mathrm{mol}$ So, The mass of the gas $\mathrm{m}=\frac{1}{2} \times 44=22 \mathrm{~g}$
BCECE-2017
Kinetic Theory of Gases
138970
Two balloons are filled one with pure, He gas and other with air respectively. If the pressure and temperature of these balloons are same, then the number of molecules per unit volume is
1 more in He gas filled balloon
2 same in both balloons
3 more in air filled balloon
4 in the ratio $1: 4$
Explanation:
B Let the volume of both balloons be same. Now, by ideal gas equation $\mathrm{PV}=\mathrm{nRT}$ If the $\mathrm{P}, \mathrm{V}$ and $\mathrm{T}$ are same the number of moles (n) will also be same whether it is He or air. $\therefore \quad \mathrm{n}=$ number of moles $\mathrm{n}=\frac{\text { Number of molecule }}{\text { Number of Avagadro }\left(\mathrm{N}_{\mathrm{a}}\right)}$ $\mathrm{n}=\frac{\text { Number of molecule }}{6.022 \times 10^{23}}$ $\mathrm{PV}=\frac{\mathrm{N}}{6 \times 10^{23}} \times \mathrm{R} \times \mathrm{T}$ $\frac{\mathrm{N}}{\mathrm{V}}=\frac{\mathrm{P} \times 6 \times 10^{23}}{\mathrm{R} \times \mathrm{T}}$ For case I- When He gas is filled- $\frac{\mathrm{N}}{\mathrm{V}}=\frac{\mathrm{P}}{\mathrm{R}} \times \frac{6 \times 10^{23}}{\mathrm{~T}}$ For case II- When the air is filled $\frac{\mathrm{N}}{\mathrm{V}}=\frac{\mathrm{P}}{\mathrm{R}} \times \frac{6 \times 10^{23}}{\mathrm{~T}}$ (Pressure is same in both case) Hence, we can see that in both case I and case II, the value of $\frac{\mathrm{N}}{\mathrm{V}}$ is same. Hence, the number of molecules per unit volume will be same in the balloons.
JCECE-2007
Kinetic Theory of Gases
138972
The equation of state for $2 \mathrm{~g}$ of oxygen at a pressure ' $P$ ' and temperature ' $T$ ', when occupying a volume ' $V$ ' will be
1 $\mathrm{PV}=\frac{1}{16} \mathrm{RT}$
2 $\mathrm{PV}=\mathrm{RT}$
3 $\mathrm{PV}=2 \mathrm{RT}$
4 $\mathrm{PV}=16 \mathrm{RT}$
Explanation:
A The equation for an ideal gas is given by $\mathrm{PV}=\mathrm{n}$ RT Now, equation of state of $2 \mathrm{~g}$ of oxygen $\because \mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}} =\frac{2}{32}$ $\mathrm{n} =\frac{1}{16}$ Putting the value of $\mathrm{n}$ in equation (i), we get $\mathrm{PV}=\frac{1}{16} \mathrm{RT}$
AIPMT 1994
Kinetic Theory of Gases
138973
The volume of a gas at $30^{\circ} \mathrm{C}$ temperature and $760 \mathrm{~mm}$ of $\mathrm{Hg}$ pressure is $100 \mathrm{cc}$. Then its volume at the same temperature and $400 \mathrm{~mm}$ of $\mathrm{Hg}$ is
138968
A balloon contains $1500 \mathrm{~m}^{3}$ of helium at $27^{\circ} \mathrm{C}$ and 4 atmospheric pressure. The volume of helium at $-3^{\circ} \mathrm{C}$ temperature and 2 atmospheric pressure will,
138969
If pressure of $\mathrm{CO}_{2}$ (real gas) in a container is given by $P=\frac{R T}{2 V-b}-\frac{a}{4 b^{2}}$, then mass of the gas in container is-
1 $11 \mathrm{~g}$
2 $22 \mathrm{~g}$
3 $33 \mathrm{~g}$
4 $44 \mathrm{~g}$
Explanation:
B Given, Pressure of $\mathrm{CO}_{2}$ (real gas) $\mathrm{P}=\frac{\mathrm{RT}}{2 \mathrm{~V}-\mathrm{b}}-\frac{\mathrm{a}}{4 \mathrm{~b}^{2}}$ We know that, The Vander waal's gas equation, $\left[\mathrm{P}+\frac{\mathrm{an}^{2}}{\mathrm{~V}^{2}}\right][\mathrm{V}-\mathrm{nb}]=\mathrm{nRT}$ $\Rightarrow \quad \mathrm{P}=\left[\frac{\mathrm{nRT}}{\mathrm{V}-\mathrm{nb}}\right]-\frac{\mathrm{n}^{2} \mathrm{a}}{\mathrm{V}^{2}}$ Comparing the equation of pressure of $\mathrm{CO}_{2}$ and Vander waal's gas equation, $\frac{\mathrm{RT}}{2 \mathrm{~V}-\mathrm{b}}-\frac{\mathrm{a}}{4 \mathrm{~b}^{2}}=\frac{\mathrm{nRT}}{\mathrm{V}-\mathrm{nb}}-\frac{\mathrm{n}^{2} \mathrm{a}}{\mathrm{V}^{2}}$ We get, $\mathrm{n}=\frac{1}{2}$ Again we know, The value of the mass $(\mathrm{m})$ $=$ molar mass $(M) \times$ Number of moles $(n)$ $\mathrm{m}=\mathrm{M} . \mathrm{n}$ Molar mass of $\mathrm{CO}_{2}=44 \mathrm{~g} / \mathrm{mol}$ So, The mass of the gas $\mathrm{m}=\frac{1}{2} \times 44=22 \mathrm{~g}$
BCECE-2017
Kinetic Theory of Gases
138970
Two balloons are filled one with pure, He gas and other with air respectively. If the pressure and temperature of these balloons are same, then the number of molecules per unit volume is
1 more in He gas filled balloon
2 same in both balloons
3 more in air filled balloon
4 in the ratio $1: 4$
Explanation:
B Let the volume of both balloons be same. Now, by ideal gas equation $\mathrm{PV}=\mathrm{nRT}$ If the $\mathrm{P}, \mathrm{V}$ and $\mathrm{T}$ are same the number of moles (n) will also be same whether it is He or air. $\therefore \quad \mathrm{n}=$ number of moles $\mathrm{n}=\frac{\text { Number of molecule }}{\text { Number of Avagadro }\left(\mathrm{N}_{\mathrm{a}}\right)}$ $\mathrm{n}=\frac{\text { Number of molecule }}{6.022 \times 10^{23}}$ $\mathrm{PV}=\frac{\mathrm{N}}{6 \times 10^{23}} \times \mathrm{R} \times \mathrm{T}$ $\frac{\mathrm{N}}{\mathrm{V}}=\frac{\mathrm{P} \times 6 \times 10^{23}}{\mathrm{R} \times \mathrm{T}}$ For case I- When He gas is filled- $\frac{\mathrm{N}}{\mathrm{V}}=\frac{\mathrm{P}}{\mathrm{R}} \times \frac{6 \times 10^{23}}{\mathrm{~T}}$ For case II- When the air is filled $\frac{\mathrm{N}}{\mathrm{V}}=\frac{\mathrm{P}}{\mathrm{R}} \times \frac{6 \times 10^{23}}{\mathrm{~T}}$ (Pressure is same in both case) Hence, we can see that in both case I and case II, the value of $\frac{\mathrm{N}}{\mathrm{V}}$ is same. Hence, the number of molecules per unit volume will be same in the balloons.
JCECE-2007
Kinetic Theory of Gases
138972
The equation of state for $2 \mathrm{~g}$ of oxygen at a pressure ' $P$ ' and temperature ' $T$ ', when occupying a volume ' $V$ ' will be
1 $\mathrm{PV}=\frac{1}{16} \mathrm{RT}$
2 $\mathrm{PV}=\mathrm{RT}$
3 $\mathrm{PV}=2 \mathrm{RT}$
4 $\mathrm{PV}=16 \mathrm{RT}$
Explanation:
A The equation for an ideal gas is given by $\mathrm{PV}=\mathrm{n}$ RT Now, equation of state of $2 \mathrm{~g}$ of oxygen $\because \mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}} =\frac{2}{32}$ $\mathrm{n} =\frac{1}{16}$ Putting the value of $\mathrm{n}$ in equation (i), we get $\mathrm{PV}=\frac{1}{16} \mathrm{RT}$
AIPMT 1994
Kinetic Theory of Gases
138973
The volume of a gas at $30^{\circ} \mathrm{C}$ temperature and $760 \mathrm{~mm}$ of $\mathrm{Hg}$ pressure is $100 \mathrm{cc}$. Then its volume at the same temperature and $400 \mathrm{~mm}$ of $\mathrm{Hg}$ is
138968
A balloon contains $1500 \mathrm{~m}^{3}$ of helium at $27^{\circ} \mathrm{C}$ and 4 atmospheric pressure. The volume of helium at $-3^{\circ} \mathrm{C}$ temperature and 2 atmospheric pressure will,
138969
If pressure of $\mathrm{CO}_{2}$ (real gas) in a container is given by $P=\frac{R T}{2 V-b}-\frac{a}{4 b^{2}}$, then mass of the gas in container is-
1 $11 \mathrm{~g}$
2 $22 \mathrm{~g}$
3 $33 \mathrm{~g}$
4 $44 \mathrm{~g}$
Explanation:
B Given, Pressure of $\mathrm{CO}_{2}$ (real gas) $\mathrm{P}=\frac{\mathrm{RT}}{2 \mathrm{~V}-\mathrm{b}}-\frac{\mathrm{a}}{4 \mathrm{~b}^{2}}$ We know that, The Vander waal's gas equation, $\left[\mathrm{P}+\frac{\mathrm{an}^{2}}{\mathrm{~V}^{2}}\right][\mathrm{V}-\mathrm{nb}]=\mathrm{nRT}$ $\Rightarrow \quad \mathrm{P}=\left[\frac{\mathrm{nRT}}{\mathrm{V}-\mathrm{nb}}\right]-\frac{\mathrm{n}^{2} \mathrm{a}}{\mathrm{V}^{2}}$ Comparing the equation of pressure of $\mathrm{CO}_{2}$ and Vander waal's gas equation, $\frac{\mathrm{RT}}{2 \mathrm{~V}-\mathrm{b}}-\frac{\mathrm{a}}{4 \mathrm{~b}^{2}}=\frac{\mathrm{nRT}}{\mathrm{V}-\mathrm{nb}}-\frac{\mathrm{n}^{2} \mathrm{a}}{\mathrm{V}^{2}}$ We get, $\mathrm{n}=\frac{1}{2}$ Again we know, The value of the mass $(\mathrm{m})$ $=$ molar mass $(M) \times$ Number of moles $(n)$ $\mathrm{m}=\mathrm{M} . \mathrm{n}$ Molar mass of $\mathrm{CO}_{2}=44 \mathrm{~g} / \mathrm{mol}$ So, The mass of the gas $\mathrm{m}=\frac{1}{2} \times 44=22 \mathrm{~g}$
BCECE-2017
Kinetic Theory of Gases
138970
Two balloons are filled one with pure, He gas and other with air respectively. If the pressure and temperature of these balloons are same, then the number of molecules per unit volume is
1 more in He gas filled balloon
2 same in both balloons
3 more in air filled balloon
4 in the ratio $1: 4$
Explanation:
B Let the volume of both balloons be same. Now, by ideal gas equation $\mathrm{PV}=\mathrm{nRT}$ If the $\mathrm{P}, \mathrm{V}$ and $\mathrm{T}$ are same the number of moles (n) will also be same whether it is He or air. $\therefore \quad \mathrm{n}=$ number of moles $\mathrm{n}=\frac{\text { Number of molecule }}{\text { Number of Avagadro }\left(\mathrm{N}_{\mathrm{a}}\right)}$ $\mathrm{n}=\frac{\text { Number of molecule }}{6.022 \times 10^{23}}$ $\mathrm{PV}=\frac{\mathrm{N}}{6 \times 10^{23}} \times \mathrm{R} \times \mathrm{T}$ $\frac{\mathrm{N}}{\mathrm{V}}=\frac{\mathrm{P} \times 6 \times 10^{23}}{\mathrm{R} \times \mathrm{T}}$ For case I- When He gas is filled- $\frac{\mathrm{N}}{\mathrm{V}}=\frac{\mathrm{P}}{\mathrm{R}} \times \frac{6 \times 10^{23}}{\mathrm{~T}}$ For case II- When the air is filled $\frac{\mathrm{N}}{\mathrm{V}}=\frac{\mathrm{P}}{\mathrm{R}} \times \frac{6 \times 10^{23}}{\mathrm{~T}}$ (Pressure is same in both case) Hence, we can see that in both case I and case II, the value of $\frac{\mathrm{N}}{\mathrm{V}}$ is same. Hence, the number of molecules per unit volume will be same in the balloons.
JCECE-2007
Kinetic Theory of Gases
138972
The equation of state for $2 \mathrm{~g}$ of oxygen at a pressure ' $P$ ' and temperature ' $T$ ', when occupying a volume ' $V$ ' will be
1 $\mathrm{PV}=\frac{1}{16} \mathrm{RT}$
2 $\mathrm{PV}=\mathrm{RT}$
3 $\mathrm{PV}=2 \mathrm{RT}$
4 $\mathrm{PV}=16 \mathrm{RT}$
Explanation:
A The equation for an ideal gas is given by $\mathrm{PV}=\mathrm{n}$ RT Now, equation of state of $2 \mathrm{~g}$ of oxygen $\because \mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}} =\frac{2}{32}$ $\mathrm{n} =\frac{1}{16}$ Putting the value of $\mathrm{n}$ in equation (i), we get $\mathrm{PV}=\frac{1}{16} \mathrm{RT}$
AIPMT 1994
Kinetic Theory of Gases
138973
The volume of a gas at $30^{\circ} \mathrm{C}$ temperature and $760 \mathrm{~mm}$ of $\mathrm{Hg}$ pressure is $100 \mathrm{cc}$. Then its volume at the same temperature and $400 \mathrm{~mm}$ of $\mathrm{Hg}$ is
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Kinetic Theory of Gases
138968
A balloon contains $1500 \mathrm{~m}^{3}$ of helium at $27^{\circ} \mathrm{C}$ and 4 atmospheric pressure. The volume of helium at $-3^{\circ} \mathrm{C}$ temperature and 2 atmospheric pressure will,
138969
If pressure of $\mathrm{CO}_{2}$ (real gas) in a container is given by $P=\frac{R T}{2 V-b}-\frac{a}{4 b^{2}}$, then mass of the gas in container is-
1 $11 \mathrm{~g}$
2 $22 \mathrm{~g}$
3 $33 \mathrm{~g}$
4 $44 \mathrm{~g}$
Explanation:
B Given, Pressure of $\mathrm{CO}_{2}$ (real gas) $\mathrm{P}=\frac{\mathrm{RT}}{2 \mathrm{~V}-\mathrm{b}}-\frac{\mathrm{a}}{4 \mathrm{~b}^{2}}$ We know that, The Vander waal's gas equation, $\left[\mathrm{P}+\frac{\mathrm{an}^{2}}{\mathrm{~V}^{2}}\right][\mathrm{V}-\mathrm{nb}]=\mathrm{nRT}$ $\Rightarrow \quad \mathrm{P}=\left[\frac{\mathrm{nRT}}{\mathrm{V}-\mathrm{nb}}\right]-\frac{\mathrm{n}^{2} \mathrm{a}}{\mathrm{V}^{2}}$ Comparing the equation of pressure of $\mathrm{CO}_{2}$ and Vander waal's gas equation, $\frac{\mathrm{RT}}{2 \mathrm{~V}-\mathrm{b}}-\frac{\mathrm{a}}{4 \mathrm{~b}^{2}}=\frac{\mathrm{nRT}}{\mathrm{V}-\mathrm{nb}}-\frac{\mathrm{n}^{2} \mathrm{a}}{\mathrm{V}^{2}}$ We get, $\mathrm{n}=\frac{1}{2}$ Again we know, The value of the mass $(\mathrm{m})$ $=$ molar mass $(M) \times$ Number of moles $(n)$ $\mathrm{m}=\mathrm{M} . \mathrm{n}$ Molar mass of $\mathrm{CO}_{2}=44 \mathrm{~g} / \mathrm{mol}$ So, The mass of the gas $\mathrm{m}=\frac{1}{2} \times 44=22 \mathrm{~g}$
BCECE-2017
Kinetic Theory of Gases
138970
Two balloons are filled one with pure, He gas and other with air respectively. If the pressure and temperature of these balloons are same, then the number of molecules per unit volume is
1 more in He gas filled balloon
2 same in both balloons
3 more in air filled balloon
4 in the ratio $1: 4$
Explanation:
B Let the volume of both balloons be same. Now, by ideal gas equation $\mathrm{PV}=\mathrm{nRT}$ If the $\mathrm{P}, \mathrm{V}$ and $\mathrm{T}$ are same the number of moles (n) will also be same whether it is He or air. $\therefore \quad \mathrm{n}=$ number of moles $\mathrm{n}=\frac{\text { Number of molecule }}{\text { Number of Avagadro }\left(\mathrm{N}_{\mathrm{a}}\right)}$ $\mathrm{n}=\frac{\text { Number of molecule }}{6.022 \times 10^{23}}$ $\mathrm{PV}=\frac{\mathrm{N}}{6 \times 10^{23}} \times \mathrm{R} \times \mathrm{T}$ $\frac{\mathrm{N}}{\mathrm{V}}=\frac{\mathrm{P} \times 6 \times 10^{23}}{\mathrm{R} \times \mathrm{T}}$ For case I- When He gas is filled- $\frac{\mathrm{N}}{\mathrm{V}}=\frac{\mathrm{P}}{\mathrm{R}} \times \frac{6 \times 10^{23}}{\mathrm{~T}}$ For case II- When the air is filled $\frac{\mathrm{N}}{\mathrm{V}}=\frac{\mathrm{P}}{\mathrm{R}} \times \frac{6 \times 10^{23}}{\mathrm{~T}}$ (Pressure is same in both case) Hence, we can see that in both case I and case II, the value of $\frac{\mathrm{N}}{\mathrm{V}}$ is same. Hence, the number of molecules per unit volume will be same in the balloons.
JCECE-2007
Kinetic Theory of Gases
138972
The equation of state for $2 \mathrm{~g}$ of oxygen at a pressure ' $P$ ' and temperature ' $T$ ', when occupying a volume ' $V$ ' will be
1 $\mathrm{PV}=\frac{1}{16} \mathrm{RT}$
2 $\mathrm{PV}=\mathrm{RT}$
3 $\mathrm{PV}=2 \mathrm{RT}$
4 $\mathrm{PV}=16 \mathrm{RT}$
Explanation:
A The equation for an ideal gas is given by $\mathrm{PV}=\mathrm{n}$ RT Now, equation of state of $2 \mathrm{~g}$ of oxygen $\because \mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}} =\frac{2}{32}$ $\mathrm{n} =\frac{1}{16}$ Putting the value of $\mathrm{n}$ in equation (i), we get $\mathrm{PV}=\frac{1}{16} \mathrm{RT}$
AIPMT 1994
Kinetic Theory of Gases
138973
The volume of a gas at $30^{\circ} \mathrm{C}$ temperature and $760 \mathrm{~mm}$ of $\mathrm{Hg}$ pressure is $100 \mathrm{cc}$. Then its volume at the same temperature and $400 \mathrm{~mm}$ of $\mathrm{Hg}$ is
138968
A balloon contains $1500 \mathrm{~m}^{3}$ of helium at $27^{\circ} \mathrm{C}$ and 4 atmospheric pressure. The volume of helium at $-3^{\circ} \mathrm{C}$ temperature and 2 atmospheric pressure will,
138969
If pressure of $\mathrm{CO}_{2}$ (real gas) in a container is given by $P=\frac{R T}{2 V-b}-\frac{a}{4 b^{2}}$, then mass of the gas in container is-
1 $11 \mathrm{~g}$
2 $22 \mathrm{~g}$
3 $33 \mathrm{~g}$
4 $44 \mathrm{~g}$
Explanation:
B Given, Pressure of $\mathrm{CO}_{2}$ (real gas) $\mathrm{P}=\frac{\mathrm{RT}}{2 \mathrm{~V}-\mathrm{b}}-\frac{\mathrm{a}}{4 \mathrm{~b}^{2}}$ We know that, The Vander waal's gas equation, $\left[\mathrm{P}+\frac{\mathrm{an}^{2}}{\mathrm{~V}^{2}}\right][\mathrm{V}-\mathrm{nb}]=\mathrm{nRT}$ $\Rightarrow \quad \mathrm{P}=\left[\frac{\mathrm{nRT}}{\mathrm{V}-\mathrm{nb}}\right]-\frac{\mathrm{n}^{2} \mathrm{a}}{\mathrm{V}^{2}}$ Comparing the equation of pressure of $\mathrm{CO}_{2}$ and Vander waal's gas equation, $\frac{\mathrm{RT}}{2 \mathrm{~V}-\mathrm{b}}-\frac{\mathrm{a}}{4 \mathrm{~b}^{2}}=\frac{\mathrm{nRT}}{\mathrm{V}-\mathrm{nb}}-\frac{\mathrm{n}^{2} \mathrm{a}}{\mathrm{V}^{2}}$ We get, $\mathrm{n}=\frac{1}{2}$ Again we know, The value of the mass $(\mathrm{m})$ $=$ molar mass $(M) \times$ Number of moles $(n)$ $\mathrm{m}=\mathrm{M} . \mathrm{n}$ Molar mass of $\mathrm{CO}_{2}=44 \mathrm{~g} / \mathrm{mol}$ So, The mass of the gas $\mathrm{m}=\frac{1}{2} \times 44=22 \mathrm{~g}$
BCECE-2017
Kinetic Theory of Gases
138970
Two balloons are filled one with pure, He gas and other with air respectively. If the pressure and temperature of these balloons are same, then the number of molecules per unit volume is
1 more in He gas filled balloon
2 same in both balloons
3 more in air filled balloon
4 in the ratio $1: 4$
Explanation:
B Let the volume of both balloons be same. Now, by ideal gas equation $\mathrm{PV}=\mathrm{nRT}$ If the $\mathrm{P}, \mathrm{V}$ and $\mathrm{T}$ are same the number of moles (n) will also be same whether it is He or air. $\therefore \quad \mathrm{n}=$ number of moles $\mathrm{n}=\frac{\text { Number of molecule }}{\text { Number of Avagadro }\left(\mathrm{N}_{\mathrm{a}}\right)}$ $\mathrm{n}=\frac{\text { Number of molecule }}{6.022 \times 10^{23}}$ $\mathrm{PV}=\frac{\mathrm{N}}{6 \times 10^{23}} \times \mathrm{R} \times \mathrm{T}$ $\frac{\mathrm{N}}{\mathrm{V}}=\frac{\mathrm{P} \times 6 \times 10^{23}}{\mathrm{R} \times \mathrm{T}}$ For case I- When He gas is filled- $\frac{\mathrm{N}}{\mathrm{V}}=\frac{\mathrm{P}}{\mathrm{R}} \times \frac{6 \times 10^{23}}{\mathrm{~T}}$ For case II- When the air is filled $\frac{\mathrm{N}}{\mathrm{V}}=\frac{\mathrm{P}}{\mathrm{R}} \times \frac{6 \times 10^{23}}{\mathrm{~T}}$ (Pressure is same in both case) Hence, we can see that in both case I and case II, the value of $\frac{\mathrm{N}}{\mathrm{V}}$ is same. Hence, the number of molecules per unit volume will be same in the balloons.
JCECE-2007
Kinetic Theory of Gases
138972
The equation of state for $2 \mathrm{~g}$ of oxygen at a pressure ' $P$ ' and temperature ' $T$ ', when occupying a volume ' $V$ ' will be
1 $\mathrm{PV}=\frac{1}{16} \mathrm{RT}$
2 $\mathrm{PV}=\mathrm{RT}$
3 $\mathrm{PV}=2 \mathrm{RT}$
4 $\mathrm{PV}=16 \mathrm{RT}$
Explanation:
A The equation for an ideal gas is given by $\mathrm{PV}=\mathrm{n}$ RT Now, equation of state of $2 \mathrm{~g}$ of oxygen $\because \mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}} =\frac{2}{32}$ $\mathrm{n} =\frac{1}{16}$ Putting the value of $\mathrm{n}$ in equation (i), we get $\mathrm{PV}=\frac{1}{16} \mathrm{RT}$
AIPMT 1994
Kinetic Theory of Gases
138973
The volume of a gas at $30^{\circ} \mathrm{C}$ temperature and $760 \mathrm{~mm}$ of $\mathrm{Hg}$ pressure is $100 \mathrm{cc}$. Then its volume at the same temperature and $400 \mathrm{~mm}$ of $\mathrm{Hg}$ is