Question

A sample of helium gas has a volume of 6.30 L at a pressure of 915 mmHg and a temperature of 27 ∘C.

please answer the following questions with the information given above.

1.) What is the pressure of the gas in atm when the volume and temperature of the gas sample are changed to 1630 mL and 349 K if the amount of gas is constant?

2.)What is the pressure of the gas in atm when the volume and temperature of the gas sample are changed to 2.33 L and 10 ∘C if the amount of gas is constant?

3.)What is the pressure of the gas in atm when the volume and temperature of the gas sample are changed to 13.0 L and 47 ∘C if the amount of gas is constant?

Answer 1

1)
Given:
Pi = 915 mmHg
Vi = 6.3 L
Vf = 1630.0 mL
= (1630.0/1000) L
= 1.63 L
Ti = 27.0 oC
= (27.0+273) K
= 300 K
Tf = 349.0 K

use:
(Pi*Vi)/(Ti) = (Pf*Vf)/(Tf)
(915 mmHg*6.3 L)/(300.0 K) = (Pf*1.63 L)/(349.0 K)
Pf = 4114 mmHg
= (4114 / 760) atm
= 5.41 atm
Answer: 5.41 atm

2)
Given:
Pi = 915 mmHg
Vi = 6.30 L
Vf = 2.33 L
Ti = 27.0 oC
= (27.0+273) K
= 300 K
Tf = 10.0 oC
= (10.0+273) K
= 283 K

use:
(Pi*Vi)/(Ti) = (Pf*Vf)/(Tf)
(915 mmHg*6.3 L)/(300.0 K) = (Pf*2.33 L)/(283.0 K)
Pf = 2334 mmHg
= 2334 / 760 atm
= 3.07 atm
Answer: 3.07 atm

3)
Given:
Pi = 915 mmHg
Vi = 6.30 L
Vf = 13.0 L
Ti = 27.0 oC
= (27.0+273) K
= 300 K
Tf = 47.0 oC
= (47.0+273) K
= 320 K

use:
(Pi*Vi)/(Ti) = (Pf*Vf)/(Tf)
(915 mmHg*6.3 L)/(300.0 K) = (Pf*13 L)/(320.0 K)
Pf = 473 mmHg
= 473/760 atm
= 0.622 atm
Answer: 0.622 atm