Question

2. A sample of gas has a pressure of 861.05 kPa at a temperature of 25°C. What temperature is required to

increase the pressure to 1985.48 kPa?​

Answer 1

The final temperature is 414.34 degrees Celsius.

Step-by-step explanation:

Gay-Lussac's law gives the relation between pressure and temperature. It states that at constant volume, pressure is directly proportional to the temperature of the gas.

`P\propto T`

We have,

Initial pressure,
`P_1=861.05\ kPa`

Initial temperature,
`T_1=25^(\circ) C= 298.15\ K`

Final pressure,
`P_2=1985.48\ kPa`

It is required to find new temperature, using Gay-Lussac's law as:

`(P_1)/(P_2)=(T_1)/(T_2)`

`T_2=(P_2T_1)/(P_1)\\\\T_2=(1985.48* 298.15)/(861.05)\\\\T_2=687.49\ K`

or

`T_2=414.34^(\circ) C`

So, the final temperature is 414.34 degrees Celsius.