Question

A sample of argon has a volume of 5.5 L and the pressure is 0.42 atm. If the final temperature is 30.° C, the final volume is 5.7 L, and the final pressure is 0.8 atm, what was the initial temperature of the argon?

Answer 1

T=15.19°C

Step-by-step explanation:

Given V1=5.5L P1=0.42atm T1=x

V2=5.7L P2=0.8atm T2=30°C

Assuming ideal gas By the law of gas

PV=nRT

Here n and R are constant throughout the process which means

`(PV)/(T)`=constant

`(P1V1)/(T1)`=
`(P2V2)/(T2)`

`(5.5*0.42)/(x)`=
`(5.7*0.8)/(30)`

on solving we get x=15.19°C