Calculate the temperature of a 969 mL sample of CH4 gas at 104 kPa with a mass of 0.647 g.?
Answer:
PV = n RT
T = PV / nR
P = 104 kPa
V = .969 L
moles = wt / mw = .647 g / (16 g/mole) = 0.0404 moles
R = 8.314 L kPa / mole °K (from here http://en.wikipedia.org/wiki/gas_constan...
So, T (in Kelvin) can be found by .
T = 104 kPa x .969 L / (.0404 moles x (8.314 L kPa/mole K))
= 300K = 27 °C
Do your own damn homework!
PS. I could give you the answer, but your professor will still want you to show your work.
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