Question

An ideal gas at `"155 kPa"` and `25^@ "C"` has an initial volume of `"1 L"`. The pressure of the ideal gas increases to `"605 kPa"` as the temperature is raised to `125^@ "C"`. What is the new volume?

Answer 1

My favorite combined gas law formula
`V_2= 0.342 L`

Explanation:

List givens, we cannot use temperature in Celsius, so convert to K, with K= C+273:
`V_1= 1 L`
`T_1=298 K`
`P_1= 155 KPa`
`V_2=?`
`P_2=605 KPa`
`T_2= 398 K`

Now Solve for `V_2`
By rearranging this formula: `(V_1P_1)/T_1=(V_2P_2)/T_2`
`V_2= (V_1P_1T_2)/(T_1P_2)`

Now just plug in numbers
`V_2= ((1 L)(155 KPa)(398 K))/((298 K)(605 KPa))`
`V_2= 0.342 L`