KINETIC THEORY OF GASES

11 )

O₂ gas is filled in a vessel. If pressure is double, temperature becomes four times, how many times its density will become.

(1/4)

(1/3)

(1/2)

(3/4)

View Solution

Answer :

PV = nRT

density = {(mass) / (Volume)} i.e.

ρ = (M / V)

∴ P × (M / ρ) = nRT

∴ P = {(nRTρ) / M}

(P / ρT) = constant

given: pressure is doubled, Temper attire is 4 times.

as (P₂ / P₁) = {(ρ₂ T₂) / (ρ₁ T₁)}

∴ {2P₁ / P₁} = (ρ₂ / ρ₁) ∙ {4T₁ / T₁}

∴ (ρ₁ / ρ₁) = (1 / 2)

∴ ρ₂ = (ρ₁ / 2)

hence density is halved

12 )

At a given volume and temperature the pressure of gas At a given volume and temperature the pressure of gas

Varies inversely as the square of its mass

Varies inversely as its mass

is independent of its mass

Varies linearly as its mass

View Solution

Answer :

PV = (M / M_o) RT

as V₁ T is constant

also R, M_o constant

∴ P ∝ M

i.e. P varies linearly as its mass.

13 )

If pressure of a gas contained in a closed vessel is increased by 0.4% when heated by 1°C the initial temperature must be.

250°C

25°C

250 K

2500 K

View Solution

Answer :

PV = nRT

as vessel is closed, V is constant.

P ∝ T

(P₂ / P₁) = (T₂ / T₁)

given: P increased by 0.4 %

P₂ = 1.04 P₁

∴ {(1.04 P₁) / P₁} = (T₂ / T₁)

∴ 1.04 = (T₂ / T₁)

T₂ = 1.04 T₁ (1)

also change in Temp = 1° C

i.e. T₁ – T₂ = 1

∴ T₁ – 1.04 T₁ = 1------- from (1)

∴ – 0.04 T₁ = 1

∴ T₁ = {(– 1) / (0.04)}

∴ T₁ = – 25° C

= 248 K

≈ 250 K

14 )

To decrease the volume of gas by 5% at constant temperature the pressure should be

Increased by 5.26%

Decreased by 5.26%

Decreased by 11%

Increased by 11%

View Solution

Answer :

PV = nRT

as Temperature is constant,

PV = constant

P₁V₁ = P₂V₂

V₂ = 95% V₁

∴ P₁ V₁ = P₂ {95 / 100} V₁

(P₂ / P₁) = (100 / 95)

∴ {(P₂ – P₁) / P₁} = (5 / 95)

∴ {(P₂ – P₁) / P₁} × 100% = (5 / 95) × 100%

i.e. % increase in pressure = 5.26%

15 )

A gas at the temperature 250 K contained in a closed vessel. If the gas is heated through 1 K, then the percentage increase in its pressure will be

0.4%

0.1%

0.8%

0.2%

View Solution

Answer :

closed vessel ⇒ volume is constant

PV = nRT

hence P ∝ T

(P₂ / P₁) = (T₂ / T₁)

∴ {(P₂ – P₁) / P₁} = {(T₂ – T₁) / T₁}

i.e. % change in pressure = {(T₂ – T₁) / T₁} × 10°

∴ % increase in P = (1 / 250) × 100

= 0.4%

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