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In Boyle's law what remains constant :

$PV$

$TV$

$\frac{V}{T}$

$\frac{P}{T}$

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Explanation

1. According to Boyle’s law PV = constant.

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Equation of gas in terms of pressure (P), absolute temperature (T) and density (d) is :

$\frac{P_{1}}{T_{1} d_{1}} = \frac{P_{2}}{T_{2} d_{2}}$

$\frac{P_{1} T_{1}}{d_{1}} = \frac{P_{2} T_{2}}{d_{2}}$

$\frac{P_{1} d_{2}}{T_{1}} = \frac{P_{2} d_{1}}{T_{1}}$

$\frac{P_{1} d_{1}}{T_{1}} = \frac{P_{2} d_{2}}{T_{2}}$

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Explanation

$1 . PV = μRT = \frac{m}{M} RT \Rightarrow P = \frac{d}{M} RT (Density d = \frac{M}{V}) \Rightarrow \frac{P}{dT} = constant or \frac{P_{1}}{d_{1} T_{1}} = \frac{P_{2}}{d_{2} T_{2}}$

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The absolute temperature of a gas is determined by

The average momentum of the molecules

The velocity of sound in the gas

The number of molecules in the gas

The mean square velocity of the molecules

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Explanation

$4 . Since v_{rms} \propto \sqrt{T} . Also mean square velocity v^{2} = v_{rms}^{2}$

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At what temperature is the root mean square velocity of gaseous hydrogen molecules is equal to that of oxygen molecules at 47°C

20 K

80 K

– 73 K

3 K

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Explanation

$1 . v_{rms} = \sqrt{\frac{3 RT}{M}} \Rightarrow T \propto M R \to constant \Rightarrow \frac{T_{H_{2}}}{T_{O_{2}}} = \frac{M_{H_{2}}}{M_{O_{2}}} \Rightarrow \frac{T_{H_{2}}}{(273 + 47)} = \frac{2}{32} \Rightarrow T_{H_{2}} = 20 K .$

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The root mean square velocity of a gas molecule of mass m at a given temperature is proportional to

$m^{0}$

$m$

$\sqrt{m}$

$\frac{1}{\sqrt{m}}$

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Explanation

$4 . v_{rms} = \sqrt{\frac{3 KT}{m}} = v_{rms} \propto \frac{1}{\sqrt{m}}$

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At room temperature, the r.m.s. speed of the molecules of certain diatomic gas is found to be 1930 m/s. The gas is :

$H_{2}$

$F_{2}$

$O_{2}$

${Cl}_{2}$

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Explanation

$1 . v_{rms} = \sqrt{\frac{3 RT}{M}} \Rightarrow M = \frac{3 RT}{v_{rms}^{2}} ∴ M = \frac{3 \times 8.3 \times 300}{{(1920)}^{2}} = 2 \times 10^{- 3} kg = 2 gm \Rightarrow Gas is hydrogen .$

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Moon has no atmosphere because :

The r.m.s. the velocity of all gases is more than the escape velocity from the moon's surface.

Its surface is not smooth.

It is quite far away from the earth.

It does not have a population and plants.

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Explanation

1. At the surface of the moon $v_{r . m . s .} > v_{escape}$ , hence molecules attain escape velocity, before

reaching their r.m.s. velocity so they escape out.

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Speed of sound in a gas is v and r.m.s. velocity of the gas molecules is c. The ratio of v to c is

$\frac{3}{γ}$

$\frac{γ}{3}$

$\sqrt{\frac{3}{γ}}$

$\sqrt{\frac{γ}{3}}$

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Explanation

$4 . We know v_{s} = \sqrt{\frac{γP}{ρ}} and v_{rms} = \sqrt{\frac{3 P}{ρ}} ∴ \frac{v_{rms}}{v_{s}} = \sqrt{\frac{γ}{3}}$

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The molecules of a given mass of a gas have a r.m.s. velocity of 200 m/sec at 27°C and $1.0 \times 10^{5} N / m^{2}$ pressure. When the temperature is 127°C and pressure is $0.5 \times 10^{5} N / m^{2},$ the r.m.s. velocity in m/sec will be

$\frac{100 \sqrt{2}}{3}$

$100 \sqrt{2}$

$\frac{400}{\sqrt{3}}$

None of the above

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Explanation

3. r.m.s. velocity doesn't depends upon pressure, it depends upon temperature only i.e.

$v_{rms} \propto \sqrt{T} \Rightarrow \frac{v_{1}}{v_{2}} = \sqrt{\frac{T_{1}}{T_{2}}} \Rightarrow \frac{200}{v_{2}} = \sqrt{\frac{(273 + 27)}{(273 + 127)}} = \sqrt{\frac{300}{400}} \Rightarrow v_{2} = \frac{400}{\sqrt{3}} m / \sec$

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Which of the following statement is true?

Absolute zero degree temperature is not zero energy temperature.

Two different gases at the same temperature pressure have equal root mean square velocities.

The root mean square speed of the molecules of different ideal gases, maintained at the same

temperature are the same.

Given sample of 1 cc of hydrogen and 1 cc of oxygen both at NTP; oxygen sample has a large

the number of molecules.

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Explanation

1. Absolute zero degree temperature is not zero energy temperature.

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