Answer to Question #349560 in General Chemistry for Siyabonga

Question

A 3.20 mol sample of a gas occupies a volume of 350 mL at 300.0 K. What is its pressure?

[1] 1.32 atm

[2] 67.0 atm

[3] 184 atm

[4] 225 atm

5] 312 atm 

Answer 1

Given:

n = 3.20 mol

V = (350 mL) × (1 L / 1000 mL) = 0.35 L

T = 300.0 K

R = 0.082 L atm mol⁻¹ K⁻¹

Solution:

Ideal gas law can be used.

Ideal gas law can be expressed as: PV = nRT

To find the pressure, solve the equation for P:

P = nRT / V

P = (3.20 mol × 0.082 L atm mol⁻¹ K⁻¹ × 300.0 K) / (0.35 L) = 224.9 atm = 225 atm

P = 225 atm

Answer: [4] 225 atm