Question

a certain mass of gas occupies 2240 volume at a pressure of 760mkhg and temperature of 320k.find it's new volume if the preddure is reduced to 700mmhg and temperature increased to 323k​

Answer 1

The new volume : V₂ = 2454.8 L

Further explanation

Given

V₁=2240 L

P₁=760 mmHg

T₁=320 K

P₂=700 mmHg

T₂=323 K

Required

Volume V₂

Solution

Combined gas law :

P₁V₁/T₁=P₂V₂/T₂

Input the value :

760 mmHg x 2240 L /320 K = 700 mmHg x V₂/323 K

V₂=(P₁V₁T₂)/P₂T₁

V₂=(760 x 2240 x 323)/(700 x 320)

V₂ = 2454.8 L