Answer to Question #310121 in General Chemistry for lubna

Question

I have an unknown volume of gas at a pressure of 50.50 KPa and a temperature of 325 K.  If I raise the pressure to 1.2 atm, decrease the temperature to 320 K, and measure the final volume to be 48 liters, what was the initial volume of the gas?

Answer 1

$P1V1T2= P2V2T2$

0.5× V1× 320= 1.2×48×325

V1= 18720/160

V1= 117L