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,

`(P_1V_1)/(T_1)=(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:

`(6atm* 18mL)/(359K)=(1atm* 300mL)/(T_2)`

`T_2=997.2K`

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