Question

A vessel of volume 8.3 x 10^-3 contains an ideal gas at temperature 27° Celsius and pressure 200kpa the gas is allowed to leak till the pressure falls 100kpa and temperature increases to 327 degree Celsius what is the amount of gas in moles will be leaked out

Answer 1

Answer:

V=8.3×10-³

T1=27°C

P1=200kPa

P2=100kPa

T2=327°C

n1=P1V/RT1 =200×10³×8.3×10-³ / 8.3×300

=2/3.

n2=P2V/RT2 =100×10³×8.3×10-³/8.3×600

=1/6.

n1-x =n2

2/3-x=1/6

x=1/2.

Explanation:

Answer 2

The amount of gas leaked out is 1.083 × 10⁻³moles.

Explanation:

The ideal gas is given by the equation:

PV = nRT

From question, the temperature and pressure changes.

The equation before leak is:

P₁V = nRT₁ ⇒ P₁/T₁ = n₁R/V ⇒ n₁ = P₁/T₁ × V/R

The equation after leak is:

P₂V = nRT₂ ⇒ P₂/T₂ = n₂R/V ⇒ n₂ = P₂/T₂ × V/R

The amount of gas leaked out is:

Δn = n₁ - n₂ = (P₁/T₁ × V/R) - (P₂/T₂ × V/R)

Δn = V/R [P₁/T₁ - P₂/T₂]

On substituting the values, we get,

Δn = (8.3 × 10⁻³)/(8.3) × [(200 × 10³)/(100 × 10₃) - (300)/(327)]

Δn = 10⁻³ × [2 - 0.917]

Δn = 10⁻³ × 1.083

∴ Δn = 1.083 × 10⁻³moles