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If density of vapours of a substance of molar mass 18 gm / mole at 1 atm pressure and 500 K is 0.36 kg m^–3 , then value of Z for the vapours is : (Take R = 0.082 L atm mole K^–1)

$\frac{41}{50}$

$\frac{50}{41}$

1.1

0.9

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Explanation

V_real = $\frac{Molar mass}{density} = \frac{18}{0 .36} .$

V_ideal = $\frac{nRT}{P} = \frac{1 \times 0 .082 \times 500}{1}$ .

So,Z = $\frac{V_{real}}{V_{ideal}} = \frac{50}{0 .082 \times 500} = \frac{50}{41}$ .

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Equal amount (mass) of methane and ethane have their total translational kinetic energy in the ratio

3: 1 then their temperatures are in the ratio.

5 : 8

45 : 8

15 : 8

8 : 5

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Explanation

E₁ = $\frac{3}{2} \times \frac{M}{30} {RT}_{1}$

E₂ = $\frac{3}{2} \times \frac{M}{30} {RT}_{2} \Rightarrow \frac{E_{1}}{E_{2}} = \frac{30}{16} \frac{T_{1}}{T_{2}} \Rightarrow \frac{3}{1} = \frac{30}{16} \frac{T_{1}}{T_{2}} \Rightarrow \frac{T_{1}}{T_{2}} = \frac{8}{5}$

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For a certain gas which deviates a little from ideal behaviour, the values of density, p

were measured at different values of pressure, P. The plot of P/p on the Y–axis versus P on the X–axis was nonlinear and had an intercept on the Y–axis, which was equal to

$\frac{RT}{M}$

$\frac{M}{RT}$

$\frac{RT}{V}$

Where M = molar mass.

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Explanation

For constant pressure V α T, hence linear dependence with the slope

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A mixture of two gases A and B in the mole ratio 2 : 3 is kept in a 2 litre vessel. A second 3 litre vessel has the same two gases in the mole ratio 3 : 5. Both gas mixtures have the same temperature and same total pressure. They are allowed to intermix ,the final temperature and total pressure are the same as the initial values, the final volume being 5 litres. Given that the molar masses are M_A and M_B, what is the mean molar mass of the final mixture?

$\frac{77 M_{A} + 123 M_{B}}{200}$

$\frac{123 M_{A} + 77 M_{B}}{200}$

$\frac{77 M_{A} + 123 M_{B}}{250}$

$\frac{{123M}_{A} + 77 M_{B}}{250}$

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At a certain temperature for which RT = 25 lit. atm. mol^–1, the density of a gas, in gm lit^–1, is d = 2.00P + 0.020 P², where P is the pressure in atmosphere. The molecular weight of the gas in gm mol^–1 is

100

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Explanation

PV = nRT

d/p = M/RT

M = 2RT

M = 2 x 25= 50

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Themolecular radius for a certain gas = 1.25 A. What is a reasonable estimate of the magnitude of the van der Wails constant, b, for the gas?

0.98 × 10^–2 litre/mole

1.43 × 10^–2 litre/mole

1.97 × 10^–2 litre/mole

3.33 × 10^–2 litre/mole

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Explanation

b = $\frac{16}{3} π r^{3} \times N_{a}$

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The critical volume of a gas is 0.072 lit. mo1^–1. The radius of the molecule will be, in cm

${(\frac{3}{4 π} \times 10^{- 23})}^{\frac{1}{3}}$

${(\frac{4 π}{3} \times 10^{- 23})}^{\frac{1}{3}}$

${(\frac{3 π}{4} \times 10^{- 23})}^{\frac{1}{3}}$

${(\frac{3}{4 π} \times 10^{- 8})}^{\frac{1}{3}}$

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Explanation

b = $\frac{16}{3} π r^{3} \times N_{a}$

0.72 × 1000 = $\frac{16}{3}$ × 6.023 × 10²³ × $π r^{3}$

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A certain gas effuses out of two different vessels A and B. A has a circular orifice while B has a square orifice of length equal to the radius of the orifice of vessel A. The ratio of rate of diffusion of the gas from vessel A to that from vessel B is

π : 1

1 : π

1 : 1

3 : 2

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Explanation

Under different conditions of pressure and temperature.

rate $α \frac{PA}{\sqrt{T . M .}}$

$∴ \frac{r_{1}}{r_{2}} = \frac{P_{1}}{P_{2}} \frac{A_{1}}{A_{2}} \sqrt{\frac{T_{2} M_{2}}{T_{1} M_{1}}}$

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A sample of a given mass of a gas at a constant temperature occupies 95cm³ under a pressure of $9 .962 \times 10^{4} N m^{- 2}$ . At the same temperature,find its volume at a pressure of $10 .13 \times 10^{4} N m^{- 2}$ is [Bihar CEE 1992]

190 cm³

93 cm³

46.5 cm³

47.5 cm³

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Explanation

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

$9 .962 \times 10^{4} \times 95 = 10 .13 \times 10^{4} \times V_{2}$

$V_{2} = 93 c m^{3}$

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A gas occupied a volume of 250 ml at 700 mm Hg pressure and 25°C. What additional pressure is required to reduce the gas volume to its 4/5^th value at the same temperature

225 mm Hg

175 mm Hg

150 mm Hg

265 mm Hg

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Explanation

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

$700 \times 250 = P_{2} \times (\frac{4}{5} \times 250)$ ; $P_{2} = 875 m m H g$

Additional pressure required = 875 – 700 = 175 mm Hg

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