Questions about Gases

Question

A weather balloon was initially at a pressure of 0.950 atm, and its volume was 35.0 L. The pressure decreased to 0.750 atm, without loss of gas or change in temperature. What was the change in the volume?

Select one:

a. decreased by 7.4 L

b. increased by 44.3 L

c. increased by 7.4 L

d. increased by 9.3 L

e. decreased by 27.6 L

Hydrogen peroxide was catalytically decomposed, and 75.3 mL of oxygen gas was collected overwater at 25°C and 742 torr. What mass of oxygen was collected? (P_water = 24 torr at 25°C)

Select one:

a. 0.0931 g

b. 0.0962 g

c. 0.00291 g

d. 0.962 g

e. 0.0993 g

Hydrochloric acid is prepared by bubbling hydrogen chloride gas through water. What is the concentration of a solution prepared by dissolving 225 L of HCl(g) at 37°C and 89.6 kPa in 5.25 L of water?

Select one:

a. 16.6  M

b. 12.5 M

c. 7.82 M

d. 1.49 M

e. 1.66 M

Calculate the root-mean-square speed of methane, CH₄ (g), at 78°C.

Select one:

a. 667 m/s

b. 350 m/s

c. 740 m/s

d. 23 m/s

e. 550 m/s

Use the van der Waals equation for real gases to calculate the pressure exerted by 1.00 mole of ammonia at 27°C in a 750-mL container. (a = 4.17 L₂·atm/mol₂, b = 0.0371 L/mol)

Select one:

a. 32.8 torr

b. 32.8 atm

c. 27.1 atm

d. 42.0 atm

e. 23.2 atm

Answer 1

When dealing with gases we are generally concerned with the following variabes:

P = pressure (usually in atm, but other units are common)

V = volume in liters

n = moles

R = gas constant (different values depending on the units of pressure)

T = temperature in Kelvin (ºC + 273 = Kelvin)

(1).

P1 = initial pressure = 0.950 atm

V1 = initial volume = 35.0 L

P2 = 0.750 atm

V2 = ?

There is no change in n (moles) and no change in T (temperature). So PV = nRT can be rearranged to

P1V1 = P2V2 and solving for V2, we have...

V2 = P1V1 / P2 = (0.950 atm)(35.0 L) / 0.750 atm

V2 = 44.3 L so it increased by 9.3 L (ANSWER d)

(2).

When a gas is collected over water, we must subtract the vapor pressure of water from the total pressure in order to obtain the pressure of the gas alone.

V = 75.3 ml = 0.0753 L

P = 742 torr - 24 torr = 718 torr x 1 atm / 760 torr = 0.945 atm

n = moles = ?

R = 0.0821 Latm/Kmol

T = 25 + 273 = 298K

PV = nRT

n = PV/RT = (0.945 atm)(0.0753 L) / (0.0821 Latm/Kmol)(298K)

n = moles of O₂ = 0.00291 mols

mass of O₂ = 0.00291 mols O2 x 32 g / mol = 0.0931 g O₂ (ANSWER a)

(3).

Concentration = moles/liter so we know the liters = 5.25 L and we need to find moles of HCl. Use PV = nRT, the ideal gas law and solve for n (moles)

n = PV/RT = (89.6 kPa)(225 L) / (8.314 L-kPa/Kmol)(310K) ... note the value of R using kPa and T is in K

n = 7.82 mols HCl

M = 7.82 mols / 5.25 L = 1.49 M (ANSWER d)

(4).

Root mean square velocity = √3RT/M

R = gas constant = 8.314 kg-m²/sec²/k-mol

T = temp in K = 78 + 273 = 351K

M = molar mass of CH4 in kg = 16 g x 1 kg/1000 g = 0.016 kg

sqrt (3)(8.314)(351)/0.016 = 740 m/s (ANSWER c)

(5). Use the van der Waals equation for real gases to calculate the pressure exerted by 1.00 mole of ammonia at 27°C in a 750-mL container. (a = 4.17 L₂·atm/mol₂, b = 0.0371 L/mol)

van der Waals equation: (P + n²a / V²) (V - nb) = nRT

P + 1²*4.17 / 0.750²) (0.750 - 1*0.0371) = (1)(0.0821)(300)

I'll leave the math to you. Solve for P