Question

A 1.2 mol sample of Kr has a volume of 867 mL. How many moles of Kr are in a 3.22 L sample at the same temperature and pressure?

A 8.90 mol sample of freon gas was placed in a balloon. Adding 3.50 mol of freon gas to the balloon increased its volume to 13.4 L. What was the initial volume of the balloon?

A mixture of He Ne, and Ar has a pressure of 20.7 atm at 28.0 °C. If the partial pressure of He is 3.08 atm and the partial pressure of Ar is 5.00 atm, what is the partial pressure of Ne?

If 3.95 molof an ideal gas has a pressure of 2.78 atm and a volume of 79.83 L, what is the temperature of the sample in degrees Celsius?

Answer 1

(1) n/V = constant , at same T and P

1.2mol /0.867L = n/3.22L

n = 4.4567mol = 4.46mol (answer)

(2)

8.90mol/V = (8.90+3.50)mol/13.4L

V = 9.618L = 9.62L. (answer)

(3)

Sum of partial pressures of each gas is equal to total pressure of the gas mixture .

3.08 + 5.00 + p_Ne = 20.7

p_Ne = 12.62 atm (answer)

(4 )

PV = nRT

T = 2.78atm×79.83L/(3.95mol×0.0821atm-L/K.mol) = 684.338K

T =684.338 - 273.15 = 411.2℃ (answer)