Question

A sample of argon has a volume of 5.0 dm³ and the pressure is 0.92 atm. If the final temperature is 30C, the final volume is 5.7 L, and the final pressure is 800 mmHg, what was the initial temperature of the argon?

Answer 1

P1=0.92atm

P2=800mmHg,760mmHg=1atm

800mmHg=800/760=1.05atm

V1=5.0dm^3=5L

V2=5.7L

T1=?

T2=30°=273+30=303K

$\frac{P1V1}{T1}=\frac{P2V2}{T2}$

$\frac{0.92atm×5L}{T1}=\frac{1.05atm×5.7L}{303K}$

$T1=232.9K$