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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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The root mean square speed of the molecules of an enclosed gas is V. What will be the root mean square speed if the pressure is doubled, the temperature remaining the same?

$\frac{v}{2}$

4 v

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Explanation

Root mean square speed is independent of pressure if the temperature remains the same.

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If the degree of freedom of gas are f, then the ratio of two specific heats $C_{P} / C_{V}$ is given by:

[MP PET 1995; BHU 1997; MP PMT 2001, 04]

$\frac{2}{f} + 1$

$1 - \frac{2}{f}$

$1 + \frac{1}{f}$

$1 - \frac{1}{f}$

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Explanation

(a)

$\frac{C_{P}}{C_{V}} = γ = 1 + \frac{2}{f}$

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The equation $(p + \frac{a}{v^{2}}) (v - b) = R T$ is known as:

Perfect gas equation

Joule Thomson's equation

Vander Waal's equation

Maxwell's equation

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Explanation

(c)

$(P + \frac{a}{V^{2}}) (V - b)$ is Vander Waal's gas equation of state.

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The temperature of an ideal gas is increased from $27 °$ to $927 ° C$ . The r.m.s. speed of its molecules becomes-

twice

half

four times

one fourth

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Explanation

(a)

$\frac{V_{r m s} (927 ° C)}{V_{r m s} (27 ° C)} = \sqrt{\frac{273 + 927}{273 + 27}}$

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An ideal gas is filled in a vessel, then

If it is placed inside a moving train, its temperature increases

Its centre of mass moves randomly

Its temperature remains constant in a moving car

None of these

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Explanation

(C)

The temperature depends on internal kinetic energy due to the random motion of gas molecules. It will not depend on the speed of the train.

Though bulk kinetic energy of gas will increase.

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If the molecular weight of two gases are $M_{1} a n d M_{2}$ , then at a temperature the ratio of root mean square velocity $v_{1} a n d v_{2}$ will be: [MP PMT 1996; AMU (Engg.) 2000; CPMT 2000; DPMT 2001]

$\sqrt{\frac{M_{1}}{M_{2}}}$

$\sqrt{\frac{M_{2}}{M_{1}}}$

$\sqrt{\frac{M_{1} + M_{2}}{M_{1} - M_{2}}}$

$\sqrt{\frac{M_{1} - M_{2}}{M_{1} + M_{2}}}$

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Explanation

(b)

$v_{r m s} = \sqrt{3 R T / M} \Rightarrow \frac{v_{1}}{v_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}$

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