A 47.0 L cylinder contains helium gas at a pressure of 23.8 atm.
Meteorologists fill a balloon with the gas in order to lift the
weather equipment into the stratosphere. What is the final pressure in
the cylinder after a 150.0 L balloon is filled to a pressure of 1.48
atm? (Note: the temperature remains constant)

Hi acmarcolini!!

Recall the Ideal Gas Law:

P.V = n.R.T 
 
where P is pressure, V is volume, n is the number of moles of gas, T
is temperature and R is the gas constant (= 0.08206 L.atm/mol.K).


If the 150.0 L balloon is filled with helium to a pressure of 1.48 atm
at a temperature T, then we have in the balloon P.V/R.T moles of
Helium:

n = P.V/R.T = 1.48*150/0.08206*T = 222/8R*T) moles


The cylinder originaly had:

n = P.V/R.T = 23.8*47/(R*T) = 1118.6/(R*T) moles

So after the ballon was filled in the cylinder remain:

(1118.6/(R*T) - 222/(R*T)) = 896.6/(R*T) moles


The final pressure in the cylinder is (T is the same in all the process):

P = n*(R*T)/V = 896.6/(R*T) * (R*T)/47 = 19.08 atm


I hope that this helps. Feel free to request for further assistance if
you find this answer unclear.


Best regards.
livioflores-ga

It's odd as I've not done one of these for 30 years

But here is (I think), the rough and ready method.

47 litres of gas at 24.8 atmospheres = 47 x 24.8 = 1118.6 litres at 1 atm
The balloon requires 150 litres at 1.8 atm = 150 x 1.8 = 222 litres at 1 atm
1118.6 - 222 = 896.6 litres left at 1 atm

to push that back into the cannister: 896.6 / 47 = 19.07659 atm