Question

The pressure of a gas in a container is 1.85 atm and occupies a volume of 12.5 L. If the original volume is reduced by half at constant temperature, what would happen to the pressure?

Answer 1

Answer:

3.7 atm i.e., pressure doubles

Explanation:

P₁ = Initial pressure = 1.85 atm

P₂ = Final pressure

V₁ = Initial volume = 12.5 L

V₂ = Final volume = 0.5V₁

T = Temperature is constant

From ideal gas law

P₁V₁ = P₂V₂

$\\\Rightarrow P_2=P_1\frac{V_1}{V_2}\\\Rightarrow P_2=1.85\frac{V_1}{0.5V_1}\\\Rightarrow P_2=1.85\frac{1}{0.5}\\\Rightarrow P_2=1.85\times 2\\\Rightarrow P_2=3.7\ atm$

∴ Final pressure is 3.7 atm i.e., pressure doubles