Question

A gas has a volume of 0.60L at 161K. Once heated, the same gas now has a volume of 14.1L at 279K and 2.44atm. What was the original pressure of the gas?

Answer 1

Answer:

31.08 atm.

Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 0.60 L

Initial temperature (T₁) = 161 K

Final volume (V₂) = 14.1 L

Final temperature (T₂) = 297 K

Final pressure (P₂) = 2.44 atm

Initial pressure (P₁) =?

The initial pressure of the gas can be obtained as follow:

P₁V₁/T₁ = P₂V₂/T₂

P₁ × 0.6 / 161 = 2.44 × 14.1 / 297

P₁ × 0.6 / 161 = 34.404 / 297

Cross multiply

P₁ × 0.6 × 297 = 161 × 34.404

P₁ × 178.2 = 5539.044

Divide both side by 178.2

P₁ = 5539.044 / 178.2

P₁ = 31.08 atm

Thus, the initial pressure of the gas was 31.08 atm.