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A gas mixture consist of 2 moles of $O_{2}$ and 4 moles of Ar at temperature T. Neglecting all vibrational modes, the total internal energy of the system is:

4RT

15RT

9RT

11RT

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Explanation

(d)

The total internal energy of system =Internal energy of oxygen molecules+Internal

The energy of oxygen molecules+Internal energy of argon molecules

$= \frac{f_{1}}{2} n 1 R T + \frac{f_{2}}{2} n_{2} R T$

= $\frac{5}{2} \times 2 R T + \frac{3}{2} \times 4 R T$

$= 11 R T$

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One mole of an ideal monatomic gas undergoes a process described by the equation $P V^{3} =$ constant. The heat capacity of the gas during this process is:

$\frac{3}{2} R$

$\frac{5}{2} R$

$2 R$

$R$

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Explanation

(d)

As we know that for the polytropic process of index $α$ ,

Specific heat capacity= $C_{v} + \frac{R}{1 - α}$

$∵ P r o c e s s, P V^{3} = c o n s t a n t \Rightarrow α = 3$

$∴ C = C_{v} + \frac{R}{1 - α} = \frac{f R}{2} + \frac{R}{1 - 3}$

where, $C_{v} = \frac{f R}{2} = \frac{3 R}{2}$

For monatomic gas, $f = 3$

$\Rightarrow C = \frac{3 R}{2} - \frac{R}{2} = R$

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A given sample of an ideal gas occupies a volume V at a pressure p and absolute temperature T. The mass of each molecule of the gas is m. Which of the following gives the density of the gas?

p/(kT)

pm / (kT)

p/ (kTV)

mkT

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Explanation

(b)

As we know that

Pressure, $p = \frac{1}{3} . \frac{n m}{V} v_{r m s}^{2}$

$∵$ nm =mass of the gas, V= volume of the gas

$∴$ $\frac{m n}{V} =$ density of the gas. Thus,

$p = \frac{1}{2} ρ v_{r m s}^{2} = \frac{1}{3} ρ \frac{3 R T}{M_{0}} (∵ v_{r m s} = \sqrt{\frac{3 R T}{M_{0}}}) ρ = \frac{p M_{0}}{R T} = \frac{p m N_{A}}{k N_{A} T} [∵ R = N_{A} k a n d M_{0} = m N_{A}] ρ = \frac{p m}{k T}$

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The molecules of a given mass of gas have r.m.s velocity of 200 ms^-1 at 27°C and 1.0 x 10⁵ Nm^-2 pressure. When the temperature and pressure of the gas are respectively, 127°C and 0.05 X 10⁵Nm^-2 , the RMS velocity of its molecules in ms^-1 is:

400/√3

100√2/3

100/3

100√2

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Explanation

(a)

$I t i s g i v e n t h a t, v_{r m s} = 200 m s^{- 1}, T_{1} = 300 K, P_{1} = 10^{5} N / m^{2} T_{2} = 400 K, P_{2} = 0.05 X 10^{5} N / m^{2} v_{r m s} = \sqrt{3 R T / m} ∴ v_{r m s} \propto \sqrt T ∴ F o r t w o d i f f e r e n t c a s e s \Rightarrow \frac{v_{r m s_{1}}}{v_{r m s_{2}}} = \sqrt{\frac{T_{1}}{T_{2}}} = \sqrt{\frac{300}{400}} v_{r m s_{2}} = \frac{400}{\sqrt{3}} m / s$

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The ratio of the specific heats C_P/C_V=γ in terms of degrees of freedom (n) is given by -

(1+1/n)

(1+n/3)

(1+2/n)

(1+n/2)

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Explanation

(c)

The specific heat of the gas at constant volume in terms of the degrees of freedom n is -

C_{V= $\frac{n}{2 R}$}

Also C_P-C_V=R

So, C_P=_{$\frac{n}{2 R}$}+R=R(1+ $\frac{n}{2}$ )

Now, γ=C_P/C_V=R(1+ $\frac{n}{2}$ )/n/2R= $\frac{2}{n}$ +1

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Two vessels separately contain two ideal gases A and B at the same temperature, the pressure of A being twice that of B. Under such conditions, the density of A is found to be 1.5 times the density of B. The ratio of molecular weight of A and B is:

2/3

3/4

1/2

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A monoatomic gas at a pressure p, having a volume V expands isothermally to a volume 2 V and then adiabatically to a volume 16 V. The final pressure of the gas is: (take γ=5/3)

64ρ

32ρ

ρ/64

16ρ

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Explanation

(c)

For isothermal expansion

ρV=ρ' x 2V [∴ V=2V]

ρ'=ρ/2

For adiabatic expansion,

$P V^{γ} = c o n s t . H e r e γ = \frac{5}{3} ∴ P' (2 V)^{\frac{5}{3}} = P_{f} (16 V)^{\frac{5}{3}} P_{f} = \frac{P}{2} {(\frac{2}{16})}^{\frac{5}{3}} = \frac{P}{64} (∵ P' = \frac{P}{2})$

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The mean free path of molecules of a gas, (radius r) is inversely proportional to :

r³

r²

√r

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Explanation

Mean free path l=1/nπr²

l ∝ 1/r2

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The molar specific heats of an ideal gas at constant pressure and volume are denoted by C_P and C_V respectively. If γ=C_P/C_V and R is the universal gas constant, then CV is equal to

1+γ/1-γ

R/(γ-1)

(γ-1)/R

γR

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Explanation

(b) The value of C_V=R/γ-1

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If $C_{p}$ and $C_{v}$ denote the specific heats (per unit mass) of an ideal gas of molecular weight M

$C_{p} - C_{v} = \frac{R}{M^{2}}$

$C_{p} - C_{v} = R$

$C_{p} - C_{v} = \frac{R}{M}$

$C_{p} - C_{v} = MR$

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Explanation

$C_{p} - C_{v} = R ∴ {MC}_{p} - {MC}_{v} = R C_{p} - C_{v} = \frac{R}{M}$

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