A gas has 1.2 L of volume, 1.8 atm of pressure and 264 K temperature. If both the volume was doubled but the pressure was halved, what would the new
temperature be?
Your answer

Temperature does not change, remains 264 K

Given

V₁= 1.2 L

P₁ = 1.8 atm

T₁ = 264

Required

The new temperature

Solution

The volume was doubled = 2V₁ = 2 x 1.2 = 2.4 L(V2)

The pressure was halved = 0.5P₁ = 0.5 x 1.8 = 0.9 atm(P2)

Combined gas law :

P₁V₁/T₁=P₂V₂/T₂

Input the value :

i believe the answer is a.