1.) The valve between a 10 L tank containing a gas at 7 atm and a 15 L tank cont

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

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