Question

the pressure on one mole of a gas at stp is doubled and temperature is raised to 546 K.what is the final volume of the gas,when one mole of gas occupies 22.4dm3 at STP

Answer 1

Given conditions are-
At STP, volume of gas is (V₁) = 22.4 dm³
Pressure is (P₁) = 1 atm
Temperature is (T₁) = 273 K
So, Volume of gas when pressure is doubled (P₂) = 2 atm
Temperature is raised to (T₂) = 546 K
Applying gas equation
P₁V₁ / T₁ = P₂V₂ / T₂
(1 X 22.4) / 273 = (2 X V₂) / 546
0.08 = 0.0036 X V₂
V₂ = 22.22 dm³