Question

A gas has a volume of 1.75L at -23°C and 150.0 kPa. At what temperature would the gas occupy 1.30L at 210.0 kPa?

Answer 1

At -13
`^(0)\textrm{C}` , the gas would occupy 1.30L at 210.0 kPa.

Step-by-step explanation:

Let's assume the gas behaves ideally.

As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-

`(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))`

where
`P_(1)` and
`P_(2)` are initial and final pressure respectively.

`V_(1)` and
`V_(2)` are initial and final volume respectively.

`T_(1)` and
`T_(2)` are initial and final temperature in kelvin scale respectively.

Here
`P_(1)=150.0kPa` ,
`V_(1)=1.75L` ,
`T_(1)=(273-23)K=250K`,
`P_(2)=210.0kPa` and
`V_(2)=1.30L`

Hence
`T_(2)=(P_(2)V_(2)T_(1))/(P_(1)V_(1))`

`\Rightarrow T_(2)=((210.0kPa)* (1.30L)* (250K))/((150.0kPa)* (1.75L))`

`\Rightarrow T_(2)=260K`

`\Rightarrow T_(2)=(260-273)^(0)\textrm{C}=-13^(0)\textrm{C}`

So at -13
`^(0)\textrm{C}` , the gas would occupy 1.30L at 210.0 kPa.