200 L of a gas at 10 atm pressure and 400 K is cooled to 100 K and reduced to 2 atm pressure. What is the new volume?

250L

Explanation:

Data;

V1 = 200L

P1 = 10atm

T1 = 400K

T2 = 100K

P2 = 2atm

V2 = ?

To solve this question, we'll have to use the combined gas equation which is a combination of Boyle's law, Charles law and pressure law.

From combined gas equation,

(P1 × V1) / T1 = (P2 × V2) / T2

P1 × V1 × T2 = P2 × V2 × T1

Solve for V2,

V2 = (P1 × V1 × T2) / P2 × T1

V2 = (10 × 200 × 100) / (2 × 400)

V2 = 200,000 / 800

V2 = 250L

The final volume of the gas is 250L