1.) The valve between a 10 L tank containing a gas at 7 atm and a 15 L tank cont
ID: 891404 • Letter: 1
Question
1.) The valve between a 10 L tank containing a gas at 7 atm and a 15 L tank containing a gas at 10 atm is opened. What is the final pressure in the tanks?
2.) Derive an expression for the gas density (m/V) by using the Ideal Gas Law and the relationship between moles of a gas (n) and its mass (m). You should substitute n=m/M where M is the molar mass of the gas and you rearrange the equation to obtain an expression for m/V.
3.) A gaseous sample of a compound has a gas density of 0.977 g/L at 710.0 torr and 100.0*C. What is the molar mass of this compound? If this compound contains only nitrogen and hydrogen and is 87.4% nitrogen by mass, what is the molecular formula of the compound?
If you could please explain each step for me, that would be so incredibly appreciated!
Explanation / Answer
1).
In tank 1 , Volume = 10 L , pressure 7 atm,
In tank 2 , volume = 15 L , pressure = Pressure = 10 atm
Calculation of number of moles in each tank by using ideal gas law
Number of moles of fist tank = pV / RT
R is the gas constant and its value is 0.08206 L atm per K per mol
= 7 atm x 10 L /RT
= 70 / RT
Number of moles of gas in second tank
n = 10.0 atm x 15 L/
= 150 / RT
Total moles = 220 / RT
= 220 / RT
We use total volume = 10 L + 15 L = 25 L
Now in that total number of moles = 220 / RT
And we can write that nRT = 220
We use ideal gas equation to find p
p = nRT / V
= 220 / 25
= 8.8 atm
So total pressure of the tank = 8.8 atm
2)
We use pV = nRT
We know n = m/ M
We put it for n
pV = (m/M )RT
by rearranging the equation we get
m/V = p M/ RT
d = pM/ RT
m / V is density
d =p M/ RT
3).
Given :
Density of gas = 0.977 g/L , p = 710.00 torr ,
T = 100 deg C
We use above density equation
d = pM/RT
M = dRT / p
Here R = 0.08206 L atm / (Kmol)
T = 100.0 deg C + 273.15 = 373.15 K
P in atm = 710.00 torr / 760 torr = 0.934 atm
Lets plug all the values
M = (0.977 g/L) x 0.08206 L atm / (Kmol) x 373.15 K / 0.934 atm
= 32.02 g/mol
M = 32.02 g/mol
Lets find empirical formula when N has 87.4 %
Lets assume total mass is 100 g then mass of N = 87.4 g
Mass of H = 12.6 g
Lets calculate moles of each
Moles of N = 87.4 g / 14 g per mol
= 6.24
Moles of H = 12.6 g / 1.0079 g per mol = 12.50
Let divide 12.50 by 6.24
Number of moles of N = 6.24 / 6.24 = 1
Number of moles of H =12.50 / 6.25 = 2
Number of N = 1 and H =2
Empirical formula is NH2
Now molecular formula
Molecular formula = n empirical formula
n = molecular formula mass / empirical formula mass
empirical formula mass = 16.0228
n =32.02 / 16.0228 = 2
molecular formula = 2 ( NH2 )
= N2H4
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.