Question

In a gas mixture, the partial pressures are argon 455 mmHg , neon 90 mmHg , and nitrogen 125 mmHg .

Part A

What is the total pressure (atm) exerted by the gas mixture?

--------------------------------------------------------------------------------------------------------------------------------------------

Part B

What is the volume, in liters, of 4.00 moles of methane gas, CH4, at 11 ∘C and 1.60 atm?

Express your answer with the appropriate units.

---------------------------------------------------------------------------------------------------------------------------------------------

A 0.211 g sample of carbon dioxide, CO2, has a volume of 549 mL and a pressure of 407 mmHg .

Part C

What is the temperature of the gas in kelvins?

Express your answer to three significant figures.

Part D

What is the temperature of the gas in degrees Celsius?

Express your answer to three significant figures.

Answer 1

A)

Total pressure = sum of all partial pressures

= 455 mmHg + 90 mmHg + 125 mmHg

= 670 mmHg

Answer: 670 mmHg

B)

Given:

P = 1.6 atm

n = 4.00 mol

T = 11.0 oC

= (11.0+273) K

= 284 K

use:

P * V = n*R*T

1.6 atm * V = 4 mol* 0.08206 atm.L/mol.K * 284 K

V = 58.3 L

Answer: 58.3 L

C)

Molar mass of CO2,

MM = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

mass(CO2)= 0.211 g

use:

number of mol of CO2,

n = mass of CO2/molar mass of CO2

=(0.211 g)/(44.01 g/mol)

= 4.794*10^-3 mol

Given:

P = 407.0 mm Hg

= (407.0/760) atm

= 0.5355 atm

V = 549.0 mL

= (549.0/1000) L

= 0.549 L

n = 0.0048 mol

use:

P * V = n*R*T

0.5355 atm * 0.549 L = 0.0048 mol* 0.08206 atm.L/mol.K * T

T = 746.4151 K

Answer: 7461 K

D)

T = 746 K

= (746-273) oC

= 473 oC

Answer: 473 oC