pressure, temp, volume, moles

Question

What is the pressure, in mmHg, of a 4.00 g sample of O2 gas, which has a temperature of 30.0 °C, and a volume of 3000 mL?

A) 0.788 mmHg

B) 1.04 mmHg

C) 78.0 mmHg

D) 7880 mmHg

E) 788 mmHg

Answer 1

Hello Miriam

Use the ideal gas law equation to solve this problem.

PV = nRT

P = ?

V = 3000 mL --> convert to L = 3000 mL x (1 L / 1000 mL) = 3 L

n = # of moles of O₂ gas --> convert 4.00 g to moles using stoichiometry

n = 4.00 g O₂ x (1 mol O₂ / 32 g O₂) = 0.125 mol O₂

R = Gas constant = 0.0821 L atm mol^-1 K^-1

T = 30.0 °C --> convert to Kelvin = 30.0 + 273.15 = 303.15 K

Rearrange the ideal gas law equation to solve for P

P = nRT / V

= (0.125 mol x 0.0821 L atm mol^-1 K^-1 x 303.15 K) / 3 L = 1.037 atm

The above answer is in unit of atm, but we need to convert it to mm Hg using the conversion factor, 1 atm = 760 mm Hg.

1.037 atm x (760 mm Hg / 1 atm) = 788 mm Hg (Answer choice E)