The next three questions pertain to the following: A 25-L tank contains 32 gm of

Question

The next three questions pertain to the following: A 25-L tank contains 32 gm of helium at 27 degree C. The molar mass of helium is 4 g/mol. (The gas constant R = 0.082 L-atm/mol-K) What is the pressure (in atm) in the tank? If half of the helium gas is withdrawn from the tank, what must be the temperature in the tank so that the pressure would remain the same? Your lungs hold 4.2 L of air at a temperature of 27 degree C and a pressure of 101.3 kPa. How many moles of air do your lungs hold? (Universal gas constant, R = 8.31 J/mol.K) A 5L tank contains 2 moles of oxygen gas, O_2, at 40 degree C. What pressure is exerted on the sides of the tank by the oxygen molecules?

Explanation / Answer

V = 25 L.
m = 32 g.
n = m/molar mass = 32/4 = 8 mol.
T = 273 + 27 = 300 K
R = 0.082 L atm/mol K

17)
Using the ideal gas equation,
PV = nRT
P = nRT/V = [8 mol x 0.082 L atm/mol K x 300 K] / 25 L.
   = 7.872 atm.

18)
P = 7.872 atm (same as before)
V = V = 25 L (same as before)
R = 0.082 L atm/mol K (same as before)
n become half of the previous value
n = 8/2 = 4 mol
Using the ideal gas equation, PV = nRT,
T = PV/nR
   = [7.872 atm x 25 L] / [4 mol x 0.082 L atm/mol K]
   = 600 K

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