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?
Answers
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)
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
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!
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.