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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?

Respuesta :

SouthernRider

Answer:

T=15.19°C

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

[tex]\frac{PV}{T}[/tex]=constant

[tex]\frac{P1V1}{T1}[/tex]=[tex]\frac{P2V2}{T2}[/tex]

[tex]\frac{5.5*0.42}{x}[/tex]=[tex]\frac{5.7*0.8}{30}[/tex]

on solving we get x=15.19°C

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