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