Question

Calculate the temperature in k of 3.05 moles of gas occupying 3.70 L at 4.12 atm

Answer 1

Answer:

${ \bf{PV = nRT}} \\ { \tt{(4.12 \times 3700) = 3.05 \times 0.083 \times T }} \\ { \tt{15244 = 0.25315 \: T}} \\ { \tt{T = 6.02 \times {10}^{4} \: kelvin }}$