Question

A gas at 155 kPa and 25 oC has an initial volume of 1.00 L. The pressure of the gas increases to 605 kPa as the temperature is raised to 125 oC. What is the new volume?

Answer 1

Answer : The new volume of gas is, 0.342 L

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 = 155 kPa

`P_2` = final pressure of gas = 605 kPa

`V_1` = initial volume of gas = 1.00 L

`V_2` = final volume of gas = ?

`T_1` = initial temperature of gas =
`25^oC=273+25=298K`

`T_2` = final temperature of gas =
`125^oC=273+125=398K`

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

`(155kPa* 1.00L)/(298K)=(605kPa* V_2)/(398K)`

`V_2=0.342L`

Therefore, the new volume of gas is, 0.342 L