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. 
    
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