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One litre of gas A and two litres of gas B, both having the same temperature 100 $°$ C and the same pressure 2.5 bar will have the ratio of kinetic energies of their molecules as:

1:1

1:2

1:4

4:1

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Explanation

$∴ K E = \frac{3}{2} k T$ , Since both the gases are at the same temperature, their mean KE are equal. The ratio is 1:1

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On 0 $°$ C, the pressure measured by the barometer is 760 mm. What will be pressure at 100 $°$ C? [AFMC 2002]

760 mm

730 mm

780 mm

None of these

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Explanation

$P \propto T \Rightarrow \frac{P_{2}}{P_{1}} = \frac{T_{2}}{T_{1}} = \frac{(273 + 100)}{(273 + 0)} = \frac{373}{273} \Rightarrow P_{2} = \frac{760 \times 373}{273} = 1038 m m$

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By what percentage, should the pressure of a given mass of gas be increased, so as to decrease its volume by 10% at a constant temperature?

7.2 %

12.5%

11.1%

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Explanation

Let $V_{1}$ be initial volume and $P_{1}$ be initial pressure.

Final volume = $V_{2} = V_{1}$ $- \frac{10}{100} V_{1} = \frac{90}{100} V_{1}$

If $P_{2}$ is final pressure, then

$P_{1} V_{1} = P_{2} V_{2}$

$\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}} = \frac{100}{90} \frac{P_{2}}{P_{1}} - 1 = \frac{10}{9} - 1 = \frac{1}{9} \frac{P_{2} - P_{1}}{P_{1}} = \frac{1}{9} \times 100 = 11.1 %$

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The ratio $\frac{C_{p}}{C_{V}} = γ$ for a gas. Its molecular weight is M. Its specification heat capacity at constant pressure is

$\frac{R}{γ - 1}$

$\frac{γ R}{γ - 1}$

$\frac{γ R}{M (γ - 1)}$

$\frac{γ R M}{(γ - 1)}$

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Explanation

$C_{P} - C_{V} = R D i v i d i n g w i t h C_{P}, w e g e t 1 - \frac{1}{γ} = \frac{R}{C_{P}} o r \frac{γ - 1}{γ} = \frac{R}{C_{P}} o r C_{P} = \frac{γ R}{γ - 1}$

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The root-mean-square velocity of the molecules in a sample of helium is $\frac{5}{7} t h$ of that of the molecules in a sample of hydrogen. If the temperature of the hydrogen gas is 0 $°$ C, that of the helium sample is about:

0 $°$ C

273 $°$ C

100 $°$ C

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Explanation

(a)

$v_{r m s} o f H e a t t ° C = \frac{5}{7} (v_{r m s} o f H_{2} a t 0 ° C) ∴ \sqrt{\frac{3 R (273 + t)}{M_{H e}}} = \frac{5}{7} \sqrt{\frac{3 R (273 + 0)}{M_{m}}} \frac{3 R (273 + t)}{4 \times 10^{- 3}} = \frac{25}{49} \frac{3 R (273)}{2 \times 10^{- 3}} t = 0 ° C$

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Four molecules have speeds 2 km/sec, 3 km/sec, 4 km/sec and 5 km/sec. The root mean square speed of these molecules (km/sec) is:

$\sqrt{\frac{54}{4}}$

$\sqrt{\frac{54}{2}}$

3.5

3 $\sqrt{3}$

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Explanation

(a)

$v_{r m s} = \sqrt{\frac{2^{2} + 3^{2} + 4^{2} + 5^{2}}{4}} = \sqrt{\frac{54}{4}} k m s^{- 1}$

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How many degrees of freedom the gas molecules have if, under STP, the gas density $ρ = 1.3 k g / m^{3}$ and the velocity of sound propagation in it is 330 m $s^{- 1}$ ?

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Explanation

(b)

$V_{s o u n d} = \sqrt{\frac{γ P}{ρ}}; 330 = \sqrt{\frac{γ \times 10^{5}}{1.3}} γ = \frac{330^{2} \times 1.3}{10^{5}} ≅ 1.3 ≅ \frac{7}{5} A l s o γ = 1 + \frac{2}{f} o r \frac{7}{5} = 1 + \frac{2}{f}; f = 5$

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The kinetic energy of one gram molecule of a gas at normal temperature and pressure is: (R = 8.31 J/mol-K)

[DPMT 1997; Pb. PMT 1997, 2000, 03; AFMC 1998; MH CET 1999]

0.56 $\times 10^{4} J$

$1.3 \times 10^{2} J$

$2.7 \times 10^{2} J$

$3.4 \times 10^{3} J$

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Explanation

(d)

Kinetic energy per g mole E = $\frac{f}{2} R T$

If nothing is said about gas then we should calculate the translational kinetic energy i.e.

$E_{T r a n s} = \frac{3}{2} R T = \frac{3}{2} \times 8.31 \times (273 + 0) = 3.4 \times 10^{3} J$

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Gases exert pressure on the walls of containing vessel because the gas molecules:

Possess momentum

collide with each other

have finite volume

obey gas laws

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Explanation

(a)

Gas molecules possess momentum, therefore, after the collision the change in momentum results. The rate of change of momentum is force and force per unit area is pressure.

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The equation of state for 5 g of oxygen at a pressure P and temperature T, when occupying a volume V, will be: (where R is the constant)

PV = 5RT

PV = $(\frac{5}{2}) R T$

PV = $(\frac{5}{16}) R T$

PV = $(\frac{5}{32}) R T$

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Explanation

(d)

Given mass of oxygen = 5g ; Pressure = P; Temperature = T; and volume = V. We know that molecular weight of oxygen = 32.

Therefore, number of moles of oxygen (n)

= $\frac{m a s s o f o x y g e n}{m o l e c u l a r m a s s o f o x y g e n} = \frac{5}{32}$

Using general gas equation that PV = nRT

We have PV = $(\frac{5}{32}) R T$ (where R = Gas constant)

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