Question

What is the final volume of a 500.0 ml gas container that increased in temperature from 299 k to 333 k while the pressure increased from 1.00 atm to 1.54 atm?

Answer 1

Use the combined gas law:

`(P _(1) V _(1) )/(T _(1) ) = (P_(2) V _(2) )/(T _(2) )`

P1 = 1.00 atm
P2 = 1.54 atm

T1 = 299 K
T2 = 333 K

T1 = 500.0 mL
T2 = ?

Solve for T2

Answer 2

Answer : The final volume will be, 361.59 ml

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 = 1 atm

`P_2` = final pressure of gas = 1.54 atm

`V_1` = initial volume of gas = 500 ml

`V_2` = final volume of gas = ?

`T_1` = initial temperature of gas = 299 K

`T_2` = final temperature of gas = 333 K

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

`(1atm* 500ml)/(299K)=(1.54atm* V_2)/(333K)`

`V_2=361.59ml`

Therefore, the final volume will be, 361.59 ml