a) What is the pressure exerted by 0.801 mol of CO2 in a 13.0 L container at 298

Question

a) What is the pressure exerted by 0.801 mol of CO2 in a 13.0 L container at 298 K? Express your answer in atm.

b) A sample of 88.0 g CO2 is held at 291 K in a 40.0 L container. What is the pressure this gas exerts on the container. Express your answer in kPa.

c) Calculate the density of nitrogen gas at a temperature of 101C and at a pressure of 1.00 atm. Express your answer in g/L.

d) A flask of fixed volume contains 1.00 mole of gaseous carbon dioxide and 88.0 g of solid carbon dioxide. The original pressure and temperature in the flask is 1.00 atm and 300. K. All of the solid carbon dioxide sublimes. The final pressure in the flask is 2.75 atm. What is the final temperature in kelvins? Assume the solid carbon dioxide takes up negligible volume.

Explanation / Answer

a) What is the pressure exerted by 0.801 mol of CO2 in a 13.0 L container at 298 K? Express your answer in atm.

PV = nRT

P = nRT/V

P = 0.801*0.082*298/13

P = 1.50563 atm

b) A sample of 88.0 g CO2 is held at 291 K in a 40.0 L container. What is the pressure this gas exerts on the container. Express your answer in kPa.

mol = mass/MW = 88/44 = 2 mol

PV = nRT

P = nRT/V

P = 2*0.082*291/40

P = 1.1931 atm

1 atm = 101.3 kPa

P = 1.1931*101.3

P = 120.86103 kPa

c) Calculate the density of nitrogen gas at a temperature of 101C and at a pressure of 1.00 atm. Express your answer in g/L.

PV = nRT

n = m/MW; D = m/V

D = P*MW/(RT)

D = 1*28/(0.082*(101+273))

D = 0.913 g/L

d) A flask of fixed volume contains 1.00 mole of gaseous carbon dioxide and 88.0 g of solid carbon dioxide. The original pressure and temperature in the flask is 1.00 atm and 300. K. All of the solid carbon dioxide sublimes. The final pressure in the flask is 2.75 atm. What is the final temperature in kelvins? Assume the solid carbon dioxide takes up negligible volume.

n = 1 CO2, m = 2 mol of CO2solid

Porinal = 1, T = 300 ;

Apply ideal gas law ratio

P1/(n1*T1) = P2/(n2*T2)

1/(1*300) = 2.75/(3*T2)

T2 = 2.75/3*300 = 275 K = 275-273 = 2 C

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