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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?

User Tsergium
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1 Answer

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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
User Peter McEvoy
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