138909
A balloon contains $500 \mathrm{~m}^{3}$ of helium at $27^{\circ} \mathrm{C}$ and 1 atmosphere pressure. The volume of the helium at $-3^{\circ} \mathrm{C}$ temperature and 0.5 atmosphere pressure will be
1 $1000 \mathrm{~m}^{3}$
2 $900 \mathrm{~m}^{3}$
3 $700 \mathrm{~m}^{3}$
4 $500 \mathrm{~m}^{3}$
Explanation:
B Given, Initial volume of balloon $=500 \mathrm{~m}^{3}$ Initial temperature $\left(\mathrm{T}_{1}\right)=27^{\circ} \mathrm{C}=300 \mathrm{~K}$ Atmospheric pressure $\left(\mathrm{P}_{1}\right)=1 \mathrm{~atm}$ Final temperature $\left(\mathrm{T}_{2}\right)=-3^{\circ} \mathrm{C}=270 \mathrm{~K}$ Final pressure $\left(\mathrm{P}_{2}\right)=0.5 \mathrm{~atm}$ We know that, Ideal gas equation $\mathrm{PV}=\mathrm{nRT}$ $\therefore \quad \frac{\mathrm{PV}}{\mathrm{T}}=$ constant Now, $\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}}$ $\frac{1 \times 500}{300}=\frac{0.5 \times \mathrm{V}_{2}}{270}$ $\mathrm{V}_{2}=\frac{5 \times 2700}{3 \times 5}$ $\mathrm{V}_{2}=900 \mathrm{~m}^{3}$
2013]
Kinetic Theory of Gases
138910
A vessel contains $16 \mathrm{~g}$ of hydrogen and $128 \mathrm{~g}$ of oxygen at standard temperature and pressure. The volume of the vessel in $\mathrm{cm}^{3}$ is:
1 $72 \times 10^{5}$
2 $32 \times 10^{5}$
3 $27 \times 10^{4}$
4 $54 \times 10^{4}$
Explanation:
C Given, mass of hydrogen $=16 \mathrm{~g}$ mass of oxygen $=128 \mathrm{~g}$ We know that, No. of mole $=\frac{\text { Mass }}{\text { Molecular weight }}$ No. of mole of $\mathrm{H}_{2}=\frac{16}{2}=8$ moles No. of mole of $\mathrm{O}_{2}=\frac{128}{32}=4$ moles $\therefore \quad$ Total moles $=8+4=12$ moles $\text { At STP, volume occupied by } 1 \text { mole }= 22.4 \mathrm{~L}$ $=22.4 \times 10^{3} \mathrm{~cm}^{3}$ $\therefore \text { Volume occupied by } 12 \text { moles } =22.4 \times 10^{3} \times 12$ $=26.88 \times 10^{4} \mathrm{~cm}^{3}$ $\therefore \text { Volume occupied by } 12 \text { moles } = 27 \times 10^{4} \mathrm{~cm}^{3}$
Shift-II
Kinetic Theory of Gases
138911
According to kinetic theory of gases,
1 A and C only
2 B and C only
3 A and B only
4 C and D only
Explanation:
B According to kinetic theory of gases, (i) The mean free path decreases on increasing the density of the molecules. $\mu =\frac{1}{\sqrt{2} \pi \mathrm{nd}^{2}}$ $\Rightarrow \quad \mu \propto \frac{1}{\mathrm{n}}$ (ii) The motion of the gas molecules freeze at $0 \mathrm{~K}$ (iii) The mean free path of gas molecules increases if temperature is increased keeping pressure constant. (iv) Average kinetic energy per molecule per degree of freedom is $\frac{1}{2} \mathrm{k}_{\mathrm{B}} \mathrm{T}$
Shift-II
Kinetic Theory of Gases
138912
For a perfect gas, two pressures $P_{1}$ and $P_{2}$ are shown in figure. The graph shows:
1 $\mathrm{P}_{1}>\mathrm{P}_{2}$
2 $\mathrm{P}_{1} \lt \mathrm{P}_{2}$
3 $\mathrm{P}_{1}=\mathrm{P}_{2}$
4 Insufficient data to draw any conclusion
Explanation:
A We know that, Ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{PdV}=\mathrm{nRdT}$ $\frac{\mathrm{dV}}{\mathrm{dT}}=\frac{\mathrm{nR}}{\mathrm{P}}$ $\Rightarrow \frac{\mathrm{dV}}{\mathrm{dT}} \propto \frac{1}{\mathrm{P}}$ $\left(\frac{\mathrm{dV}}{\mathrm{dT}}\right)_{2}>\left(\frac{\mathrm{dV}}{\mathrm{dT}}\right)_{1}$ $\Rightarrow \mathrm{P}_{1}>\mathrm{P}_{2}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Kinetic Theory of Gases
138909
A balloon contains $500 \mathrm{~m}^{3}$ of helium at $27^{\circ} \mathrm{C}$ and 1 atmosphere pressure. The volume of the helium at $-3^{\circ} \mathrm{C}$ temperature and 0.5 atmosphere pressure will be
1 $1000 \mathrm{~m}^{3}$
2 $900 \mathrm{~m}^{3}$
3 $700 \mathrm{~m}^{3}$
4 $500 \mathrm{~m}^{3}$
Explanation:
B Given, Initial volume of balloon $=500 \mathrm{~m}^{3}$ Initial temperature $\left(\mathrm{T}_{1}\right)=27^{\circ} \mathrm{C}=300 \mathrm{~K}$ Atmospheric pressure $\left(\mathrm{P}_{1}\right)=1 \mathrm{~atm}$ Final temperature $\left(\mathrm{T}_{2}\right)=-3^{\circ} \mathrm{C}=270 \mathrm{~K}$ Final pressure $\left(\mathrm{P}_{2}\right)=0.5 \mathrm{~atm}$ We know that, Ideal gas equation $\mathrm{PV}=\mathrm{nRT}$ $\therefore \quad \frac{\mathrm{PV}}{\mathrm{T}}=$ constant Now, $\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}}$ $\frac{1 \times 500}{300}=\frac{0.5 \times \mathrm{V}_{2}}{270}$ $\mathrm{V}_{2}=\frac{5 \times 2700}{3 \times 5}$ $\mathrm{V}_{2}=900 \mathrm{~m}^{3}$
2013]
Kinetic Theory of Gases
138910
A vessel contains $16 \mathrm{~g}$ of hydrogen and $128 \mathrm{~g}$ of oxygen at standard temperature and pressure. The volume of the vessel in $\mathrm{cm}^{3}$ is:
1 $72 \times 10^{5}$
2 $32 \times 10^{5}$
3 $27 \times 10^{4}$
4 $54 \times 10^{4}$
Explanation:
C Given, mass of hydrogen $=16 \mathrm{~g}$ mass of oxygen $=128 \mathrm{~g}$ We know that, No. of mole $=\frac{\text { Mass }}{\text { Molecular weight }}$ No. of mole of $\mathrm{H}_{2}=\frac{16}{2}=8$ moles No. of mole of $\mathrm{O}_{2}=\frac{128}{32}=4$ moles $\therefore \quad$ Total moles $=8+4=12$ moles $\text { At STP, volume occupied by } 1 \text { mole }= 22.4 \mathrm{~L}$ $=22.4 \times 10^{3} \mathrm{~cm}^{3}$ $\therefore \text { Volume occupied by } 12 \text { moles } =22.4 \times 10^{3} \times 12$ $=26.88 \times 10^{4} \mathrm{~cm}^{3}$ $\therefore \text { Volume occupied by } 12 \text { moles } = 27 \times 10^{4} \mathrm{~cm}^{3}$
Shift-II
Kinetic Theory of Gases
138911
According to kinetic theory of gases,
1 A and C only
2 B and C only
3 A and B only
4 C and D only
Explanation:
B According to kinetic theory of gases, (i) The mean free path decreases on increasing the density of the molecules. $\mu =\frac{1}{\sqrt{2} \pi \mathrm{nd}^{2}}$ $\Rightarrow \quad \mu \propto \frac{1}{\mathrm{n}}$ (ii) The motion of the gas molecules freeze at $0 \mathrm{~K}$ (iii) The mean free path of gas molecules increases if temperature is increased keeping pressure constant. (iv) Average kinetic energy per molecule per degree of freedom is $\frac{1}{2} \mathrm{k}_{\mathrm{B}} \mathrm{T}$
Shift-II
Kinetic Theory of Gases
138912
For a perfect gas, two pressures $P_{1}$ and $P_{2}$ are shown in figure. The graph shows:
1 $\mathrm{P}_{1}>\mathrm{P}_{2}$
2 $\mathrm{P}_{1} \lt \mathrm{P}_{2}$
3 $\mathrm{P}_{1}=\mathrm{P}_{2}$
4 Insufficient data to draw any conclusion
Explanation:
A We know that, Ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{PdV}=\mathrm{nRdT}$ $\frac{\mathrm{dV}}{\mathrm{dT}}=\frac{\mathrm{nR}}{\mathrm{P}}$ $\Rightarrow \frac{\mathrm{dV}}{\mathrm{dT}} \propto \frac{1}{\mathrm{P}}$ $\left(\frac{\mathrm{dV}}{\mathrm{dT}}\right)_{2}>\left(\frac{\mathrm{dV}}{\mathrm{dT}}\right)_{1}$ $\Rightarrow \mathrm{P}_{1}>\mathrm{P}_{2}$
138909
A balloon contains $500 \mathrm{~m}^{3}$ of helium at $27^{\circ} \mathrm{C}$ and 1 atmosphere pressure. The volume of the helium at $-3^{\circ} \mathrm{C}$ temperature and 0.5 atmosphere pressure will be
1 $1000 \mathrm{~m}^{3}$
2 $900 \mathrm{~m}^{3}$
3 $700 \mathrm{~m}^{3}$
4 $500 \mathrm{~m}^{3}$
Explanation:
B Given, Initial volume of balloon $=500 \mathrm{~m}^{3}$ Initial temperature $\left(\mathrm{T}_{1}\right)=27^{\circ} \mathrm{C}=300 \mathrm{~K}$ Atmospheric pressure $\left(\mathrm{P}_{1}\right)=1 \mathrm{~atm}$ Final temperature $\left(\mathrm{T}_{2}\right)=-3^{\circ} \mathrm{C}=270 \mathrm{~K}$ Final pressure $\left(\mathrm{P}_{2}\right)=0.5 \mathrm{~atm}$ We know that, Ideal gas equation $\mathrm{PV}=\mathrm{nRT}$ $\therefore \quad \frac{\mathrm{PV}}{\mathrm{T}}=$ constant Now, $\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}}$ $\frac{1 \times 500}{300}=\frac{0.5 \times \mathrm{V}_{2}}{270}$ $\mathrm{V}_{2}=\frac{5 \times 2700}{3 \times 5}$ $\mathrm{V}_{2}=900 \mathrm{~m}^{3}$
2013]
Kinetic Theory of Gases
138910
A vessel contains $16 \mathrm{~g}$ of hydrogen and $128 \mathrm{~g}$ of oxygen at standard temperature and pressure. The volume of the vessel in $\mathrm{cm}^{3}$ is:
1 $72 \times 10^{5}$
2 $32 \times 10^{5}$
3 $27 \times 10^{4}$
4 $54 \times 10^{4}$
Explanation:
C Given, mass of hydrogen $=16 \mathrm{~g}$ mass of oxygen $=128 \mathrm{~g}$ We know that, No. of mole $=\frac{\text { Mass }}{\text { Molecular weight }}$ No. of mole of $\mathrm{H}_{2}=\frac{16}{2}=8$ moles No. of mole of $\mathrm{O}_{2}=\frac{128}{32}=4$ moles $\therefore \quad$ Total moles $=8+4=12$ moles $\text { At STP, volume occupied by } 1 \text { mole }= 22.4 \mathrm{~L}$ $=22.4 \times 10^{3} \mathrm{~cm}^{3}$ $\therefore \text { Volume occupied by } 12 \text { moles } =22.4 \times 10^{3} \times 12$ $=26.88 \times 10^{4} \mathrm{~cm}^{3}$ $\therefore \text { Volume occupied by } 12 \text { moles } = 27 \times 10^{4} \mathrm{~cm}^{3}$
Shift-II
Kinetic Theory of Gases
138911
According to kinetic theory of gases,
1 A and C only
2 B and C only
3 A and B only
4 C and D only
Explanation:
B According to kinetic theory of gases, (i) The mean free path decreases on increasing the density of the molecules. $\mu =\frac{1}{\sqrt{2} \pi \mathrm{nd}^{2}}$ $\Rightarrow \quad \mu \propto \frac{1}{\mathrm{n}}$ (ii) The motion of the gas molecules freeze at $0 \mathrm{~K}$ (iii) The mean free path of gas molecules increases if temperature is increased keeping pressure constant. (iv) Average kinetic energy per molecule per degree of freedom is $\frac{1}{2} \mathrm{k}_{\mathrm{B}} \mathrm{T}$
Shift-II
Kinetic Theory of Gases
138912
For a perfect gas, two pressures $P_{1}$ and $P_{2}$ are shown in figure. The graph shows:
1 $\mathrm{P}_{1}>\mathrm{P}_{2}$
2 $\mathrm{P}_{1} \lt \mathrm{P}_{2}$
3 $\mathrm{P}_{1}=\mathrm{P}_{2}$
4 Insufficient data to draw any conclusion
Explanation:
A We know that, Ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{PdV}=\mathrm{nRdT}$ $\frac{\mathrm{dV}}{\mathrm{dT}}=\frac{\mathrm{nR}}{\mathrm{P}}$ $\Rightarrow \frac{\mathrm{dV}}{\mathrm{dT}} \propto \frac{1}{\mathrm{P}}$ $\left(\frac{\mathrm{dV}}{\mathrm{dT}}\right)_{2}>\left(\frac{\mathrm{dV}}{\mathrm{dT}}\right)_{1}$ $\Rightarrow \mathrm{P}_{1}>\mathrm{P}_{2}$
138909
A balloon contains $500 \mathrm{~m}^{3}$ of helium at $27^{\circ} \mathrm{C}$ and 1 atmosphere pressure. The volume of the helium at $-3^{\circ} \mathrm{C}$ temperature and 0.5 atmosphere pressure will be
1 $1000 \mathrm{~m}^{3}$
2 $900 \mathrm{~m}^{3}$
3 $700 \mathrm{~m}^{3}$
4 $500 \mathrm{~m}^{3}$
Explanation:
B Given, Initial volume of balloon $=500 \mathrm{~m}^{3}$ Initial temperature $\left(\mathrm{T}_{1}\right)=27^{\circ} \mathrm{C}=300 \mathrm{~K}$ Atmospheric pressure $\left(\mathrm{P}_{1}\right)=1 \mathrm{~atm}$ Final temperature $\left(\mathrm{T}_{2}\right)=-3^{\circ} \mathrm{C}=270 \mathrm{~K}$ Final pressure $\left(\mathrm{P}_{2}\right)=0.5 \mathrm{~atm}$ We know that, Ideal gas equation $\mathrm{PV}=\mathrm{nRT}$ $\therefore \quad \frac{\mathrm{PV}}{\mathrm{T}}=$ constant Now, $\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}}$ $\frac{1 \times 500}{300}=\frac{0.5 \times \mathrm{V}_{2}}{270}$ $\mathrm{V}_{2}=\frac{5 \times 2700}{3 \times 5}$ $\mathrm{V}_{2}=900 \mathrm{~m}^{3}$
2013]
Kinetic Theory of Gases
138910
A vessel contains $16 \mathrm{~g}$ of hydrogen and $128 \mathrm{~g}$ of oxygen at standard temperature and pressure. The volume of the vessel in $\mathrm{cm}^{3}$ is:
1 $72 \times 10^{5}$
2 $32 \times 10^{5}$
3 $27 \times 10^{4}$
4 $54 \times 10^{4}$
Explanation:
C Given, mass of hydrogen $=16 \mathrm{~g}$ mass of oxygen $=128 \mathrm{~g}$ We know that, No. of mole $=\frac{\text { Mass }}{\text { Molecular weight }}$ No. of mole of $\mathrm{H}_{2}=\frac{16}{2}=8$ moles No. of mole of $\mathrm{O}_{2}=\frac{128}{32}=4$ moles $\therefore \quad$ Total moles $=8+4=12$ moles $\text { At STP, volume occupied by } 1 \text { mole }= 22.4 \mathrm{~L}$ $=22.4 \times 10^{3} \mathrm{~cm}^{3}$ $\therefore \text { Volume occupied by } 12 \text { moles } =22.4 \times 10^{3} \times 12$ $=26.88 \times 10^{4} \mathrm{~cm}^{3}$ $\therefore \text { Volume occupied by } 12 \text { moles } = 27 \times 10^{4} \mathrm{~cm}^{3}$
Shift-II
Kinetic Theory of Gases
138911
According to kinetic theory of gases,
1 A and C only
2 B and C only
3 A and B only
4 C and D only
Explanation:
B According to kinetic theory of gases, (i) The mean free path decreases on increasing the density of the molecules. $\mu =\frac{1}{\sqrt{2} \pi \mathrm{nd}^{2}}$ $\Rightarrow \quad \mu \propto \frac{1}{\mathrm{n}}$ (ii) The motion of the gas molecules freeze at $0 \mathrm{~K}$ (iii) The mean free path of gas molecules increases if temperature is increased keeping pressure constant. (iv) Average kinetic energy per molecule per degree of freedom is $\frac{1}{2} \mathrm{k}_{\mathrm{B}} \mathrm{T}$
Shift-II
Kinetic Theory of Gases
138912
For a perfect gas, two pressures $P_{1}$ and $P_{2}$ are shown in figure. The graph shows:
1 $\mathrm{P}_{1}>\mathrm{P}_{2}$
2 $\mathrm{P}_{1} \lt \mathrm{P}_{2}$
3 $\mathrm{P}_{1}=\mathrm{P}_{2}$
4 Insufficient data to draw any conclusion
Explanation:
A We know that, Ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{PdV}=\mathrm{nRdT}$ $\frac{\mathrm{dV}}{\mathrm{dT}}=\frac{\mathrm{nR}}{\mathrm{P}}$ $\Rightarrow \frac{\mathrm{dV}}{\mathrm{dT}} \propto \frac{1}{\mathrm{P}}$ $\left(\frac{\mathrm{dV}}{\mathrm{dT}}\right)_{2}>\left(\frac{\mathrm{dV}}{\mathrm{dT}}\right)_{1}$ $\Rightarrow \mathrm{P}_{1}>\mathrm{P}_{2}$
