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0.850 moles of N2 originally at 85°C is cooled such that it now occupies 17.55L at 1.25 atm. What is the final temperature of the gas?

Respuesta :

you know at condition 2
P = 1.25atm
V = 17.55L
T = ?
R = 0.08206 Latm/molK
n = 0.850 mol

so you can use
PV = nRT
to solve for T

T = PV / (nR)
T = (1.25atm) x (17.55L) / (0.850mol x 0.08206 Latm/molK)
T = 314.5K = 41.1°C

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