Question

A gas has a volume of 3.25 liters at 54 C and 231 kPa of pressure. At what temperature will the same gas take up 4.35 liters of space and have a pressure of 168 kPa?

Answer 1

Answer: 318 K

Step-by-step 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 = 231 kPa

`P_2` = final pressure of gas = 168 kPa

`V_1` = initial volume of gas = 3.25 L

`V_2` = final volume of gas = 4.35 L

`T_1` = initial temperature of gas =
`54^oC=273+54=327K`

`T_2` = final temperature of gas = ?

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

`(231* 3.25)/(327)=(168* 4.35)/(T_2)`

`T_2=318K`

At 318 K of temperature will the same gas take up 4.35 liters of space and have a pressure of 168 kPa