Question

A 0.250 L sample of chlorine gas is collected at a pressure of 105.3 kPa and a temperature of 505 K. What will the volume be if the pressure changes to 100.3 kPa and the temperature decreases to 495 K?

Answer 1

V2 = 0.257 L

Step-by-step explanation:

T1 = 505 K

P1 = 105.3 kPa = 1.04 atm

V1 = 0.250 L

T2 = 495 K

P2 = 100.3 kPa = 0.990 atm

V2 = ?

P1V1/T1 = P2V2/T2 must be changed to T2P1V1/T1/P2

495 x 1.04 x 0.250 ÷ 505 ÷ 0.990 = V2

V2 = 0.257 L

So 0.257 L is the answer!!