What is the temperature of 0.5 moles of water vapor that occupies 120 dm3 and applies a pressure of 15,000 pa to its container

Use the ideal gas law:

PV = nRT

so, T = PV / nR

n=0.5
V= 120 dm^3 = 120 L  (1 dm^3 = 1 L)
R = 1/12
P = 15,000 Pa =  0.147 atm     (1 pa = 9.86 10^{-6} )

Put the values:

T = PV / nR
T = (0.147) (120) / (0.5) (1/12)
T= 426 K

Ithink 2 is the correct