Question

​29. A gas has a volume of 1.75 L at -23°C and 150.0 kPa.

At what temperature would the gas occupy 1.30 L at
210.0 kPa?

Answer 1

The answer for the following mention bellow.

• Therefore the final temperature of the gas is 260 k

Step-by-step explanation:

Given:

Initial pressure (
`P_(1)`) = 150.0 kPa

Final pressure (
`P_(2)`) = 210.0 kPa

Initial volume (
`V_(1)`) = 1.75 L

Final volume (
`V_(2)`) = 1.30 L

Initial temperature (
`T_(1)`) = -23°C = 250 k

To find:

Final temperature (
`T_(2)`)

We know;

According to the ideal gas equation;

P × V = n × R ×T

where;

P represents the pressure of the gas

V represents the volume of the gas

n represents the no of moles of the gas

R represents the universal gas constant

T represents the temperature of the gas

We know;

`(P*V)/(T)` = constant

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

Where;

(
`P_(1)`) represents the initial pressure of the gas

(
`P_(2)`) represents the final pressure of the gas

(
`V_(1)`) represents the initial volume of the gas

(
`V_(2)`) represents the final volume of the gas

(
`T_(1)`) represents the initial temperature of the gas

(
`T_(2)`) represents the final temperature of the gas

So;

`(150 * 1.75)/(210 * 1.30)` =
`(260)/(T_(2) )`

(
`T_(2)`) =260 k

Therefore the final temperature of the gas is 260 k