Question

the Temperature of a sample of an ideal gas in a sealed 5.0 l container is raised from 27 oC to 77oC. If the initial pressure of the gas was 3.0 atm, what is the final pressure?

Answer 1

Answer:
The pressure increases to 3.5 atm.

Solution:
According to Gay-Lussac's Law, " At constant volume and mass the pressure of gas is directly proportional to the applied temperature".

For initial and final states of a gas the equation is,

P₁ / T₁ = P₂ / T₂
Solving for P₂,
P₂ = P₁ T₂ / T₁ ----- (1)

Data Given;
P₁ = 3 atm

T₁ = 27 °C + 273 = 300 K

T₂ = 77 °C + 273 = 350 K

Putting values in eq. 1,

P₂ = (3 atm × 350 K) ÷ 300 K

P₂ = 3.5 atm