# What is the new Kelvin temperature of a gas if originally 62.0 ml of the gas was at a temperature of 373 K and a pressure of 120 KPA and the volume changes to 120.0 ml. and the pressure changes to...

What is the new Kelvin temperature of a gas if originally 62.0 ml of the gas was at a temperature of 373 K and a pressure of 120 KPA and the volume changes to 120.0 ml. and the pressure changes to 80.0 KPa.

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We can use the combine gas law to solve this problem. It is expressed as:

`(P_1 V_1)/(T_1) = (P_2 V_2)/(T_2)`

• P1 = 120kPa = 1.1844 atm
• V1 = 62 mL = 0.062L
• T1 = 373
• P2 = 80 kPa = 0.7895 atm
• V2 = 120 mL = 0.12L
• T2 = ?

For pressure, we convert the unit from kPa to atm and for the volume form mL to L.

1 kPa = 0.00987atm
1 L = 1000 mL

Substitute the given values and solve algebraically.

`(P_1 V_1)/(T_1) = (P_2 V_2)/(T_2)`

`(1.1844 *0.062)/(373) = (0.7895* 0.12)/(T_2)`

`4.1969x10^(-4) = (0.09474)/(T_2)`

`T_2 =0.09474/(4.1969x10^(-4))`

`T_2 = 481 K` -> answer

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