Question

A sample of oxygen has a pressure of 5.00 x 10^2 kPa

at 40.00° C. What will be the temperature in Kelvin at a
pressure of 6.30 x 10² kPa?
• Use -273.15°C for absolute zero.
• Round the answer to three significant figures.

Answer 1

Answer:394.569K

Step-by-step explanation:

Gay-Lussac's law states that at constant volume,the pressure of a gas is directly proportional to its temperature.

The pressure,temperature at any states with constant volume can be related as
`(P_(1))/(T_(1)) =(P_(2))/(T_(2))`

where
`(P_(1),T_(1))` are pressure and temperature at initial state and
`(P_(2),T_(2))` are pressure and temperature at final state.

Given,
`P_(1)=5* 10^(2)`
`kPa`

`T_(1)=313.15K`

Given,
`P_(2)=6.3* 10^(2)`
`kPa`

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

`(5* 10^(2))/(313.15)=(6.3* 10^(2))/(T_(2))`

So,
`T_(2)=(6.3* 10^(2)* 313.15)/(5* 10^(2))`

`T_(2)=394.569K`