Question

A sample of helium gas has a volume of 6.70 L at a pressure of 845 mmHgand a temperature of 26 ∘C . What is the final pressure of the gas, in atmospheres, when the volume and temperature of the gas sample are changed to the following, if the amount of gas does not change?

1550 mL and 315 K

P1 =		atm

Part B

2.45 L and 18 ∘C

P2 =		atm

Part C

14.0 L and 42 ∘C

P3 =		atm

Answer 1

P1V1/T1 = P2V2 / T2

P1 = 845 mmHg = 1.11 atm

V 1 = 6.70 L

T1 = 26⁰C = 299 K

A) P2 = ? , T2 = 315 K , V2 = 1550 mL = 1.55 L

P2 = (P1V1 T2 / T1 V2)

P2 = (1.11 x 6.70 x 315 / 299 x 1.55)

P2 = ( 2342.655 / 463.45)

P2 = 5.05 atm

B) P2 = ? , T2 = 18⁰C = 291 K = V2 = 2.45 L

P2 = (P1V1 T2 / T1 V2)

P2 = (1.11 x 6.70 x 291 / 299 x 2.45)

P2 = 2164.167 / 732.55

P2 = 2.95 atm

C) P2 = ? , T2 = 42⁰C = 315 K , V2 = 14.0 L

P2 = (P1V1 T2 / T1 V2)

P2 = (1.11 x 6.70 x 315 / 299 x 14)

P2 = 2342.655 / 4186

P2 = 0.56 atm