What is the density of #NH_3# gas at 435 K and 1.00 atm?

1 Answer
Feb 16, 2017

Answer:

From the ideal gas equation: #n/V=(P)/(RT)#. Agreed? And we assume ideality for ammonia gas (good enuff for a first approx.).
We get #rho~~0.5*g*L^-1#

Explanation:

But #n="Mass"/"Molar mass"#.

Thus #"Mass"/"Molar mass"=(PV)/(RT)#

And on rearrangement, #"Mass"/"V"=(P)/(RT)xx"Molar mass"#

But, by definition, #"density, "rho="Mass"/"Volume"=(P)/(RT)xx"Molar mass"#

#=(1.00*cancel(atm)xx17.03*g*cancel(mol^-1))/(0.0821*L*cancel(atm)*cancel(K^-1)*cancel(mol^-1)xx435*cancelK)=#

#??g*L^-1#

So this expression gives us an answer in #g*L^-1# as required for a density.