Question

A 14.0 gram sample of oxygen gas has a volume of 809 milliliters at a pressure of 3.72 atm. The temperature of the O₂ gas sample is ________ °C.

Answer 1

PV = nRT

where, P = pressure = 3.72 atm

V = volume = 809 mL = 0.809 L

n = number of moles = 14.0 g / 32.0 g/mol = 0.438 mol

R = Gas constant

T = temperature = ?

3.72 * 0.809 = 0.438* 0.0821* T

3.01 = 0.0360 * T

T = 3.01 / 0.0360 = 83.6 K = -189.4 ^oC

Therefore, temperature = -189.4 ^oC