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Physics MCQs for NEET — Practice Questions with Answers

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20 L of SO₂ diffuses through a porous partition in 60s. Volume of O₂ diffuse under similar condition in 30s will be:-

12.14 L

14.14 L

18.14 L

28.14 L

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Explanation

$\frac{V_{S O_{2}} / t_{S O_{2}}}{V_{O_{2}} / t_{O_{2}}} = \sqrt{\frac{M_{O_{2}}}{M_{S O_{2}}}}$

= $\frac{\frac{20}{60}}{\frac{V_{O_{2}}}{30}} = \sqrt{\frac{32}{64}}$

= $\frac{10}{V_{O_{2}}} = \frac{1}{\sqrt{2}}$

$V_{O_{2}} = 10 \sqrt{2}$ = 14.14 L

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In vander wall's equation of state for a non-ideal gas, which term accounts for intermolecular forces is:-

v-b

(P+a/V²)

(RT)^-1

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Explanation

(P+a/V²)(V-b) = RT

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Which of the following can be most readily liquafied?

(Given : Value of 'a' for NH₃ = 4.17, CO₂ = 3.59, SO₂ = 6.71, Cl₂ = 6.49)

NH₃

Cl₂

SO₂

CO₂

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Explanation

Higher the value of 'a' more the gas liquifiable. 'a' indicates ATTRACTIVE FORCE.

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At moderate pressure, the Vander wall's equation become:-

PV_m = RT

P(V_m-b) = RT

(P+a/V²_m) (V_m )= RT

P = RT/V_m + a/V²_m

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Explanation

at low pressure, V high

(P+a/V²_m) (V_m )= RT

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The compressibility factor for a real gas at high pressure is:-

1+Pb/RT

1-Pb/RT

1-RT/Pb

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Explanation

At high 'P'

(P) (V-b) =RT

PV-Pb = RT

PV/RT - Pb/RT =1

PV/RT=1+Pb/RT

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A gas diffuses four time quickly as oxygen. The molecular weight of gas is:

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Explanation

r ∝ 1/ $\sqrt{M w}$

r_o2/r_gas $\sqrt{\frac{(M w) g a s}{32}}$

1/4 = $\sqrt{\frac{(M w) g a s}{32}}$

1/16 = Mw/32

Mw = 2

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Which of the following mixtures of gases at room temperature does not obey Dalton's law of partial pressure?

He and O₂

NH₃ and HCL

H₂ and He

N₂ and He

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Explanation

Non-reacting gases follow Datton's law of partial pressure.

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Rate of diffusion of LPG (a mixture of n-butane and propane) is 1.25 times of SO₃. Hence, mass fraction of n-butane in LPG is

0.75

0.25

0.50

0.67

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Explanation

Use, $\frac{r_{1}}{r_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}$

_{$r_{LPG} / r_{{SO}_{3}}$} = √(80/ M_LPG) = 1.25
Therefore, 80/M = (1.25)2
Therefore, M_LPG = 51.2
M = (M₁(butane ) X₁ + M₂(propane)X₂) / (X₁ + X₂)
51.2 = 58X₁ + 44(1-X₁ ) / 1
X₁ ≈ 0.50

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An open flask contining air is heated from 300 K to 500 K. What percentage of air will be escaped to the atmosphere, if pressure is keeping constant?

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Explanation

For open contianer, P and v are constant. Suppose initial moles = n

1/n₁T₁ = 1/n₂T₂

1/nx300 = 1/n₂ x 500

n₂ = (3/5)n

Escape moles = n-(3/5)n = (2/5)n

% of escape air = (2/5)n/n x 100 =40

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A 40 mL of a mixture of H₂and O₂was placed in a gas burette at 18^oC and 1 atm pressure. A spark was supplied so that the formation of water was complete . The remaining pure gas had a volume of 10 mL at 18^oC and 1 atm pressure. If the remaining gas was H₂, what was the initial mole percentage of H₂ in the mixture?

25%

50%

35%

75%

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Explanation

$H_{2} + \frac{1}{2} O_{2} \to H_{2} O Volume before reaction a b 0 Volume after reaction a - 2 b 0 b Now, a + b = 40 and, a - 2 b = 10 On solving the above two equations we get, a = 30 mL and b = 10 mL Now, mole fraction of H_{2} = \frac{a}{a + b} \frac{30}{30 + 10} = 0.75$

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