Question

Part A) A weather balloon is inflated to a volume of 29.8 L at a pressure of 741 mmHg and a temperature of 24.0 ∘C. The balloon rises in the atmosphere to an altitude where the pressure is 390. mmHg and the temperature is -13.9 ∘C. Assuming the balloon can freely expand, calculate the volume of the balloon at this altitude.

Part B) What is the temperature of 0.47 mol of gas at a pressure of 1.3 atm and a volume of 12.7 L ? Express the temperature in kelvins to two significant figures.

Part C) What is the pressure in a 11.0-L cylinder filled with 35.1 g of oxygen gas at a temperature of 345 K ? Express your answer to three significant figures with the appropriate units.

Part D)A sample of N2O gas has a density of 3.00 g/L at 298 K.What must be the pressure of the gas (in mmHg)?

Answer 1

A)

Given:

Pi = 741 mmHg

Pf = 390 mmHg

Vi = 29.8 L

Ti = 24.0 oC

= (24.0+273) K

= 297 K

Tf = -13.9 oC

= (-13.9+273) K

= 259.1 K

use:

(Pi*Vi)/(Ti) = (Pf*Vf)/(Tf)

(741 mmHg*29.8 L)/(297.0 K) = (390 mmHg*Vf)/(259.1 K)

Vf = 49.4 L

Answer: 49.4 L

B)

Given:

P = 1.3 atm

V = 12.7 L

n = 0.47 mol

use:

P * V = n*R*T

1.3 atm * 12.7 L = 0.47 mol* 0.08206 atm.L/mol.K * T

T = 428 K

Answer: 428 K

C)

Molar mass of O2 = 32 g/mol

mass(O2)= 35.1 g

use:

number of mol of O2,

n = mass of O2/molar mass of O2

=(35.1 g)/(32 g/mol)

= 1.097 mol

Given:

V = 11.0 L

n = 1.0969 mol

T = 345.0 K

use:

P * V = n*R*T

P * 11 L = 1.0969 mol* 0.08206 atm.L/mol.K * 345 K

P = 2.8231 atm

Answer: 2.82 atm

D)

Let volume be 1 L and mass be 3.00 g

Molar mass of N2O,

MM = 2*MM(N) + 1*MM(O)

= 2*14.01 + 1*16.0

= 44.02 g/mol

mass(N2O)= 3.00 g

use:

number of mol of N2O,

n = mass of N2O/molar mass of N2O

=(3 g)/(44.02 g/mol)

= 6.815*10^-2 mol

Given:

V = 1.0 L

n = 0.0682 mol

T = 298.0 K

use:

P * V = n*R*T

P * 1 L = 0.0682 mol* 0.08206 atm.L/mol.K * 298 K

P = 1.6678 atm

= 1.6678 * 760 mmHg

= 1268 mmHg

Answer: 1268 mmHg