Question

How many moles of oxygen must be placed
in a 3.00 L container to exert a pressure of 2.00 atm at 25.0°C?

Formula: PV = nRT(R = 0.0821 Lmr002-1.jpgatm/molmr002-2.jpgK)

Which variables are given?

pressure

volume

temperature

moles

Answer 1

Use the ideal gas law: PV = nRT → n = PV / (RT).

Given:
- P = 2.00 atm
- V = 3.00 L
- T = 25.0°C = 298.15 K
- R = 0.0821 L·atm/(mol·K)

Compute:
n = (2.00 atm)(3.00 L) / (0.0821 L·atm/(mol·K) · 298.15 K)
≈ 6.00 / 24.48
≈ 0.245 mol

So 0.245 moles of O2 are required. Variables given: pressure, volume, and temperature (moles is the unknown).