Question

A gas has a volume of 3.25 liters at 54 C and 231 kPa of pressure. At what temperature will the same gas take up 4.35 liters of space and have a pressure of 168 kPa?

Answer 1

Answer: 318 K

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 231 kPa

$P_2$ = final pressure of gas = 168 kPa

$V_1$ = initial volume of gas = 3.25 L

$V_2$ = final volume of gas = 4.35 L

$T_1$ = initial temperature of gas = $54^oC=273+54=327K$

$T_2$ = final temperature of gas = ?

Now put all the given values in the above equation, we get:

$\frac{231\times 3.25}{327}=\frac{168\times 4.35}{T_2}$

$T_2=318K$

At 318 K of temperature will the same gas take up 4.35 liters of space and have a pressure of 168 kPa