KINETIC%20THEORY%20OF%20GASES

1 )

Volume, pressure and temperature of an ideal gas are V, P, T respectively. If mass of molecule is m, then its density is [ k_B = Boltzmann's constant]

Select correcct option from the following options

(A) [P / (k_BT)] (B) [Pm / (k_BT)]

View Solution

Answer :

PV = nRT --- ideal gas equation

as n = (M / M_o) where M is mass & M_o is molar mass then

∴ PV = {(MRT) / (M_o)}

∴ {(PM_o) / (RT)} = (M / v)

as {(mass) / (volume)} = (M / v) = density = δ hence

∴ ρ = {(PM_o) / (RT)}

∴ρ = {(pm × NA) / (RT)}

where M_o = m × NA, m is mass of molecule

NA is Avogadro's number

ρ = [(pm) / (RT / NA)]

as (R / NA) = k_B = Boltzmann's constant

ρ = [(Pm) / (k_BT)]

2 )

The temperature of an ideal gas at atmospheric pressure is 300 K and volume 1 m³. If temperature and volume become double, then pressure will be

Select correct option from the following options

(A) 4 × 10⁵ Nm^–2 (B) 2 × 10⁵ Nm^–2

View Solution

Answer :

PV = nRT

∴ (PV / T) = nR = constant

∴ {(P₁V₁) / T₁} = {(P₂ V₂) / T₂}

∴ P₂ = {(P₁ V₁ T₂) / (T₁ V₂)}

Give: T₂ = 2T₁

V₂ = 2V₁

∴ P₂ {(P₁ V₁ × 2T₁) / (T₁ × 2V₁)}

P₂ = P₁ = 1 atm

hence P₂ = 1 × 10⁵ Nm^–2

3 )

At 100 K and 0.1 atmospheric pressure, the volume helium gas is 10 liters. If volume and pressure are doubled, its temperature will change to

127 K

400 K

25 K

200 K

View Solution

Answer :

PV = nRT

i.e. PV ∝ T.

give: P, V both doubled,

as PV ∝ T, temperature will become 4 times of original hence Temp. will

change to 4 × 100 = 400K

4 )

What is the mass of 2 liters of nitrogen at 22.4 atmospheric pressure and 273 K. (R = 8.314 Jmol k^–1)

66 g

56 g

46 g

36 g

View Solution

Answer :

V = 2 lire = 2 × 10^–3 m³

P = 22.4 atm = 22.4 × 1.013 × 10⁵ = 22.6 × 10⁵ N/m²

T = 273 K

As PV = nRT

(PV / RT) = n

∴ n = no. of moles = [{22.6 × 10⁵ × 2 × 10^–3} / {8.314 × 273}] = 2

as n = (m / m_o) = {(mass of gas) / (molars mass of gas)}

2 = (m / 28) - - - m_o = 28 for N₂

∴ m = 56 gm

5 )

An electron tube was sealed off during manufacture at a pressure of 1 × 10^–7 torr at 27°C. Its volume is 100 cm³. The number of molecules that remain in the tube is_______(density of mercury is 13.6 gcm^–3) (1 torr = 133 Pa)

(A) 3.9 × 10¹¹ (B) 3 × 10¹⁶

View Solution

Answer :

given: P = 1.2 × 10^–7 torr = 1.2 × 10^–7 × 133 -------- as 1 torr = 133 Pa

= 159.6 × 10^–7 Pa = 159.6 × 10^–7 N/m²

V = 100 cm³ = 100 × 10^–6 m³

T = 27°C = 300k

As PV = NK_BT

∴ N = No of molecules = {(PV) / (K_BT)}

= {(159.6 × 10^–7 × 100 × 10^–6) / (1.38 × 10^–23 × 300)}

= 0.385 × 10⁺¹²

= 3.85 × 10¹¹

Topic Categories