Question

Determine the temperature in Celsius of 1.40 moles of gas contained in a 2.000-L vessel at a pressure of 6.340 atm.

Answer 1

Ideal gas law

PV = nRT

P = 6.340 atm

V = 2.0 L

n = 1.40 mol

R = 0.082058 L atm/K mol

$T = \frac{PV}{nR} \\ T = \frac{6.340 \times 2.0}{1.40 \times 0.082058} = 110.37 \;K$

T = 110.37 – 273.15 = -162.78 ºC

Answer: -162.78 ºC.