Question

A gas has a volume of 1.8Lat−26◦Cand 147 kPa. At what temperature would the gas occupy 1.33 L at 217 kPa?

Answer 1

Answer: At temperature of 269 K the gas would occupy 1.33 L at 217 kPa

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 = 147 kPa

`P_2` = final pressure of gas = 217 kPa

`V_1` = initial volume of gas = 1.8 L

`V_2` = final volume of gas = 1.33 L

`T_1` = initial temperature of gas =
`-26^oC=273-26=247K`

`T_2` = final temperature of gas = ?

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

`(147kPa* 1.8L)/(247K)=(217kPa* 1.33L)/(T_1)`

`T_2=269K`

Thus at 269 K temperature the gas would occupy 1.33 L at 217 kPa