Answer to Question #116091 in General Chemistry for natty

Question

A sample of argon has a volume of 1.2 L at STP. If the temperature is increased to 22°C and the pressure is lowered to 0.80 atm, what will the new volume be, in L?

Answer 1

P₁ = 1.0 atm;

P₂ = 0.80 atm;

T₁ = 0 °C = 273.15 K;

T₂ = 22 °C = 295.15 K;

V₁ = 1.2 L;

V₂ = unknown;

Solution:

(P₁×V₁)/(T₁) = (P₂×V₂)/(T₂);

V₂ = (P₁×V₁×T₂)/(T₁×P₂);

V₂ = (1.0 atm × 1.2 L × 295.15 K)/(273.15 K×0.80 atm) = 1.62 L;

Answer: 1.62 L