Question

At what temperature would a gas have a volume of 13.5 L at a pressure of 0.723 atm, if it had a volume of 17.8 L at a pressure of 0.612 atm and a temperature of 28 C

Answer 1

Answer : The initial temperature of gas would be, 269.7 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 = 0.723 atm

$P_2$ = final pressure of gas = 0.612 atm

$V_1$ = initial volume of gas = 13.5 L

$V_2$ = final volume of gas = 17.8 L

$T_1$ = initial temperature of gas = ?

$T_2$ = final temperature of gas = $28^oC=273+28=301K$

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

$\frac{0.723atm\times 13.5L}{T_1}=\frac{0.612atm\times 17.8L}{301K}$

$T_1=269.7K$

Thus, the initial temperature of gas would be, 269.7 K