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.

Respuesta :

Answer:394.569K

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 [tex]\frac{P_{1}}{T_{1}} =\frac{P_{2}}{T_{2}}[/tex]

where [tex](P_{1},T_{1})[/tex] are pressure and temperature at initial state and [tex](P_{2},T_{2})[/tex] are pressure and temperature at final state.

Given,[tex]P_{1}=5\times 10^{2}[/tex][tex]kPa[/tex]

[tex]T_{1}=313.15K[/tex]

Given,[tex]P_{2}=6.3\times 10^{2}[/tex][tex]kPa[/tex]

[tex]\frac{P_{1}}{T_{1}} =\frac{P_{2}}{T_{2}}[/tex]

[tex]\frac{5\times 10^{2}}{313.15}=\frac{6.3\times 10^{2}}{T_{2}}[/tex]

So,[tex]T_{2}=\frac{6.3\times 10^{2}\times 313.15}{5\times 10^{2}}[/tex]

[tex]T_{2}=394.569K[/tex]