Question

A sample of argon gas has a volume of 785 mL at a pressure of 1.36 atm and a temperature of 118 ∘C.

Part A

What is the volume of the gas in milliliters when the pressure and temperature of the gas sample are changed to 634 mmHg and 327 K , if the amount of gas remains the same?

V1 =

_____ mL

Part B

What is the volume of the gas in milliliters when the pressure and temperature of the gas sample are changed to 0.850 atm and 70 ∘C, if the amount of gas remains the same?

V2=___mL

Part C

What is the volume of the gas in milliliters when the pressure and temperature of the gas sample are changed to 15.2 atm and -15 ∘C, if the amount of gas remains the same?

V3=___mL

Answer 1

A)

Given:

Pi = 1.36 atm

Pf = 634.0 mm Hg

= (634.0/760) atm

= 0.8342 atm

Vi = 785 mL

Ti = 118.0 oC

= (118.0+273) K

= 391 K

Tf = 327.0 K

use:

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

(1.36 atm*785.0 mL)/(391.0 K) = (0.8342 atm*Vf)/(327.0 K)

Vf = 1070 mL

Answer: 1070 mL

B)

Given:

Pi = 1.36 atm

Pf = 0.850 atm

Vi = 785 mL

Ti = 118.0 oC

= (118.0+273) K

= 391 K

Tf = 70.0 oC

= (70.0+273) K

= 343 K

use:

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

(1.36 atm*785 mL)/(391.0 K) = (0.85 atm*Vf)/(343.0 K)

Vf = 1102 mL

Answer: 1.10*10^3 mL

C)

Given:

Pi = 1.36 atm

Pf = 15.2 atm

Vi = 785 mL

Ti = 118.0 oC

= (118.0+273) K

= 391 K

Tf = -15.0 oC

= (-15.0+273) K

= 258 K

use:

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

(1.36 atm*785 mL)/(391.0 K) = (15.2 atm*Vf)/(258.0 K)

Vf = 46.3 mL

Answer: 46.3 mL