Determine the temperature of 1.42 mol of a gas contained in a 3.00-L container at a pressure of 123 kPa

Chemistry

1 answer:

Rashid [163]3 years ago

6 0

Answer:

The temperature is 30,92K

Explanation:

We use the formula PV=nRT. We convert the unit of pressure in kPa into atm.

101,325kPa----1atm

121kPa-------x=(121,3kPax 1 atm)/101,325kPa=1, 2 atm

PV=nRT---->T= (PV)/(RT)

T=(1,2 atm x 3L)/(1,42 mol x 0,082 l atm/K mol )= 30, 91721058 K

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