Asked by Thomas

If 63.1 g of NH3 occupies 35.1 L under a pressure of 93.1 in. Hg, what is the temperature of the gas, in °C?
Answered by oobleck
PV=nRT, so since n and R are constant here, PV/T is constant.
1atm = 29.9 in Hg, and 1 mole occupies 22.4L at STP, so you want T such that
(93.1)(35.1)/(T+273.15) = (29.9)(63.1/17 * 22.4)/(273.15)
Answered by DrBob222
The following appears to me to be a much simpler way to solve the problem. Stick with PV = nRT
P = 93.1 inches x 1 atm/29.9 inches = 3.12 atm
V = 35.1 L
T= ?
n = 63.1g/17 = 3.72
R = 0.0821
3.12*35.1 = 3.72*0.0821*T
T = 359 K. Then 359- 273 = 86 C
Answered by oobleck
As usual, DrBob222 cuts to the chase.
Luckily, my little calculation agrees with his!
Answered by DrBob222
Yes, the answers agree. I checked that FIRST THING because I know oobleck is always right.
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