Question

A 18 mL sample of gas has a temperature of 86 ˚C and a pressure of 6 atm. What temperature would be needed for the same amount of gas to fit into a 300 mL flask at standard pressure? (at constant moles)

Answer 1

Answer : The temperature needed for the same amount of gas would be, 997.2 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 = 6 atm

$P_2$ = final pressure of gas = 1 atm

$V_1$ = initial volume of gas = 18 mL

$V_2$ = final volume of gas = 300 mL

$T_1$ = initial temperature of gas = $86^oC=273+86=359K$

$T_2$ = final temperature of gas = ?

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

$\frac{6atm\times 18mL}{359K}=\frac{1atm\times 300mL}{T_2}$

$T_2=997.2K$

Thus, the temperature needed for the same amount of gas would be, 997.2 K