Question

A sample of oxygen at 44◦C occupies 373 mL.

If this sample later occupies 1102 mL at 60◦C

and 1.8 atm, what was its original pressure?

Answer in units of atm

Answer 1

T₁=44 °C=317 K

V₁=373 ml=0.000373 m³

T₂=60°C=333 K

V₂=1102 ml=0.001102 m³

P₂=1.8 atm=182385 Pa

R=8.31 J/(K mole)

P₁ -?

$PV=nRT$

$n=\frac{P_2V_2}{RT_2}=\frac{182385x0.001102}{8.31x333}=0.0726 mole$

$P_1=\frac{nRT_1}{V_1}=\frac{0.0726x8.31x317}{0.000373}=512729 Pa$