Answer to Question #298702 in Chemistry for abc

Question

A gas has a volume of 3.00 L and a pressure of 75.4 kPa. when the volume expands to 4.00 L and the pressure drops to 72.7 kPa, the gas temperature is 0C. what was the initial temperature of the gas?

Answer 1

V₁ = 3.00 L

P₁ = 75.4 kPa

T₁ = unknown

V₂ = 4.00 L

P₂ = 72.7 kPa

T₂ = 0°C = 273.15 K

Solution:

The Combined Gas Law can be used:

P₁V₁/T₁ = P₂V₂/T₂

Cross-multiply to clear the fractions:

P₁V₁T₂ = P₂V₂T₁

Divide to isolate T₁:

T₁ = (P₁V₁T₂) / (P₂V₂)

Plug in the numbers and solve for T₁:

T₁ = (75.4 kPa × 3.00 L × 273.15 K) / (72.7 kPa × 4.00 L) = 212.47 K

T₁ = 212.47 K or −60.7°C

Answer: The initial temperature of the gas is 212.47 K (or −60.7°C)