Question

Q. 0.40 mol of a monoatomic gas fills a 1 dm^3 container to a pressure 1.013×10^6 pa. It is expanded reversibally and adiabatically until a pressure of 1.013×10^5 pa is reached, calculate:
1. What is the final volume of gas?
2. initial and final temperature of gas?
3. work done by gas during expansion?​

Answer 1

Given info : 0.40 mol of a monoatomic gas fills a 1 dm³ container to a pressure 1.013 × 10⁶ pa. It is expanded reversibally and adiabatically until a pressure of 1.013 × 10⁵ pa is reached.

To find :

1. what is the final volume of gas ?
2. initial and final temperature of gas ?
3. work done by the gas during expansion ?

solution : using formula, P1V1^γ = P2V2^γ

here, P1 = 1.013 × 10⁶ Pa , V1 = 1 dm³ , V2 = ? P2 = 1.013 × 10⁵ Pa

for monoatomic, γ = 1.66

so, (1.013 × 10⁶) × (1)^(1.66) = (1.013 × 10⁵) × (V2)^(1.66)

⇒V2 = (10)^(1/1.66) dm³ = 3.98 ≈ 4 dm³

therefore volume of the gas is 4 dm³

initial temperature,

PV = nRT

⇒1.013 × 10⁶ Pa × 1 dm³ = 0.4 mol × 25/3 J/mol/K × T

⇒1.013 × 10⁶ N/m² × 10¯³ m³ = 0.4 mol × 25/3 J/mol/K × T

⇒T = 304 K

so the initial temperature is 304 K

final temperature is found by, Tf = Ti(Vi/Vf)^γ-1

= 304(1/4)^(1.66 - 1)

= 120.6 K

work done by the gas, W = (P2V2 - P1V1)/(1 - γ)

= (1.013 × 10⁵ Pa × 4 × 10¯³ m³ - 1.013 × 10⁶ × 1 × 10¯³ m³)/(1 - 1.66)

= 6 × 1.013 × 10²/0.66

= 921 J