If a 2.8 mole sample of gas initially occupies 54 L at 3.8 atm, what is its temperature in K?
n=

R=

T=

893.7K

Explanation:

From the question, we obtained the following:

V = 54L

n = 2.8moles

P = 3.8 atm

R = 0.082atm.L/mol /K

T =?

Applying the ideal gas equation, we have:

PV = nRT

T = PV /nR

T = (3.8x54)/(2.8x0.082)

T = 893.7K

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