Question

A sample of helium gas has a volume of 5.10 L at a pressure of 815 mmHg and a temperature of 26 ∘C.

A) What is the pressure of the gas in atm when the volume and temperature of the gas sample are changed to 1610 mLand 338 K if the amount of gas is constant?

B) 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 11 ∘C if the amount of gas is constant?

C)What is the pressure of the gas in atm when the volume and temperature of the gas sample are changed to 12.4 L and 44 ∘C if the amount of gas is constant?

Answer 1

A)

Given:

Pi = 815 mmHg

Vi = 5.1 L

Vf = 1610 mL

= (1610/1000) L

= 1.61 L

Ti = 26.0 oC

= (26.0+273) K

= 299 K

Tf = 338.0 K

use:

(Pi*Vi)/(Ti) = (Pf*Vf)/(Tf)

(815 mmHg*5.1 L)/(299.0 K) = (Pf*1.61 L)/(338.0 K)

Pf = 2918 mmHg

= 2918/760 atm

= 3.84 atm

Answer: 3.84 atm

B)

Given:

Pi = 815 mmHg

Vi = 5.10 L

Vf = 2.33 L

Ti = 26.0 oC

= (26.0+273) K

= 299 K

Tf = 11.0 oC

= (11.0+273) K

= 284 K

use:

(Pi*Vi)/(Ti) = (Pf*Vf)/(Tf)

(815 mmHg*5.1 L)/(299.0 K) = (Pf*2.33 L)/(284.0 K)

Pf = 1694 mmHg

= 1694/760 atm

= 2.23 atm

Answer: 2.23 atm

C)

Given:

Pi = 815 mmHg

Vi = 5.10 L

Vf = 12.4 L

Ti = 26.0 oC

= (26.0+273) K

= 299 K

Tf = 44.0 oC

= (44.0+273) K

= 317 K

use:

(Pi*Vi)/(Ti) = (Pf*Vf)/(Tf)

(815 mmHg*5.1 L)/(299.0 K) = (Pf*12.4 L)/(317.0 K)

Pf = 355.4 mmHg

= 355.4/760 atm

= 0.468 atm

Answer: 0.468 atm