Question

A sample of a gas has a pressure of 1.26 bar at 58.0∘

c. What is the pressure of the gas at 118∘C?

Answer 1

According to Gay-Lussac's law of pressure–temperature: pressure (P) of gas is directly proportional to temperature (T).

The formula is given by:

`(P)/(T)=k (constant))` or,

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

`P_(1)` (Pressure of gas) =1.26 bar

`T_(1)` (temperature of gas) =
`58.0^(o)C`

`P_(2)` (Pressure of gas) =?

`T_(2)` (temperature of gas) =
`118.0^(o)C`

Substitute the above values in formula (1)

`(1.26 bar)/(58.0^(o)C)=(P_(2))/(118.0^(o)C)`

`(1.26 bar)/(58.0^(o)C)=(P_(2))/(118.0^(o)C)`

Convert
`^(o)C` into kelvin

`(1.26 bar)/((58.0^(o)C +273 ))* (118.0^(o)C+273 )= P_(2)`

`P_(2) =(1.26 bar)/(331 K)* (391 K)`

=
`1.488 bar \simeq 1.49 bar`

Thus, pressure of the gas at
`118.0^(o)C` is 1.49 bar.