Solved problems

1. A gas at 87^oC and 0.62 atm occupies a volume of 0.452 liters. Calculate the volume of the gas at standard conditions (i.e., at 0 ^OC e 1 atm) and its number of moles.

Remembering that P₁V₁/T₁ = P₂V₂/T₂ we have that V₂ = (P₁V₁T₂)/(T₁P₂), from which we can calculate the volume at 1 atm and 0 ^oC (273 K).

V2 = 76.5/360 = 0.212 liters

n = PV / RT = (.452)0.62/(0.00820)(360) = 0,0094

Alternatively remembering that the volume of 1 mole at standard conditions is 22.41 liters

n = 0.212/22.41 = 0,0094

2. An ideal gas is contained in a tank (1) of unknown volume at the pressure of 1 atm. By opening a valve, the gas is allowed to expand in another vacuum container (2) having a volume of 0.5 liters. Once reached the equilibrium a pressure of 530 mm Hg is measured. Assuming an isotherm process, calculate the volume of tank 1.

Below are summarized the data of the problem:

P1 = 1 atm = 760 mm

V1 = unknown

P2 = 530 mm

V2 = 0.5 + V1

According to Boyle' law, at constant temperature, P₁V₁ = P₂V₂ and thus

P1V1 = P2V2 = 760 V1 = 530 (V1+ 0.5)	760 V1 = 530 V1 + (530)(0.5)	760 V1 = 530 V1 + 265
760 V1 - 530 V1 = 265	230 V1 = 265	V1 = 265/ 230 = 1.15 liters

3. 0.896 grams of a gaseous compound, containing only nitrogen (atomic weight = 14) and oxygen (atomic weight = 16), are contained in a tank of 524 cm³ at 730 mm Hg and 28 ^oC. Calculate the molecular weight and the molecular formula of the compound. The value for R is 0.082 liters atm mole^-1 K^-1.

First we calculate the number of moles by using the equation of ideal gases

Since R is given in liters atm mole^-1 K^-1, some transformation are needed to apply the equation above.

Pressure. 1 atm is 760 mm Hg then 730 mm correspond 730/760 = 0.96 atm.

Temperature. K = ^oC + 273 then T = 28 + 273 = 301 K

Volume. 524 cm³  correspond to 0.524 litri

Therefore n = PV/RT = (0.96) (0.524)/(301) (0.082) = 0.503/24.69 = 0.0203 moli.

Using the given number of grams and remembering that moles = grams/molecular weight, we can calculate the molecular weight

mw = grams/moles = 0.896/0.0203 = 44

According to the problem the compound contains only nitrogen and oxygen and indicating with x and y the moles of N and O respectively it should be

mw = x(N) + y(O) = 14x + 16y = 44

It easy to verify that this equation is verified for x = 2 (nitrogen) and y = 1 (oxygen) and thus the molecular formula is N₂O.

4. The valve connecting two tanks each containing a different gas, is opened to allow the diffusion of the two gases. Assuming that:

Tank a ==> V = 5 liters, initial pressure = 9 atm

Tank b ==> V = 10 litri, initial pressure = 6 atm

Calculate the final pressure if the diffusion is accomplished at constant temperature.

Rembering that at constant temperature P_iV_i = P_fV_f we have

Gas a

inizial condition Pi = 9; Vi = 5

final condition Pf = x; Vf = 5 + 10 = 15

PiVi = PfVf; (9) (5) = (Pf)(15)

from which P_f = 3.

from which P_f = 4

Therefore, according to Dalton's law, P_t = 4 + 3 = 7.

5. 2.69 grams of PCl₅ (Mw = 208) are contained in a one liter flask at 250 ^oC and 1 atm. Considering that the gas can dissociate according to the equation PCl_5(g) = PCl_3(g) + Cl_2(g), verify if in the given condition of temperature and pressure the  gas is dissociated. In the case the dissociation takes place, calculate the partial pressure of the three gases.

Assume R = 0.082 liters atm mole^-1 K^-1.

As first let's calculate the moles of PCl_5, n = grams/Mw = 2.69/208 = 0.013 moles. In absence of dissociation, according to the equation of state of gases, n = (PV)/(RT), we should obtain n = 0.013 moles. However, if we calculate n using the given value for P,V and T we get:

n = (PV)/(RT) = (1 atm)(1 liter)/(0.082)(250 + 273) = 1/42.8 = 0,0233 moles.

Since 0.0233 is greater than the number of moles calculated for undissociated PCl₅ then we can conclude that dissociation occured.

According to the dissociation equation, for each mole of PCl₅ dissociated there is the formation of 1 mole of Cl₂ and 1 mole di PCl₃ then moles Cl₂ = moles di PCl₃. Indicating with x the fraction of  PCl₅ dissociated and by 2x the total moles of products formed, we can write

total moles = 0.0233 = (moles PCl_{5 initial} - x ) + 2 x = 0.013 + x

x = 0.0233 - 0.013 = 0.0103

Thus at the equilibrium are presents 0.0103 moles of Cl₂ e 0.0103 moles of PCl₃ from which we can calculate the partial pressures:

partial pressure for Cl₂ is P = (nRT)/V= (0.0103)(0.082)(523) = 0.441 atm

partial pressure for PCl₃ is, of coures, the same calculated for Cl₂

partial pressure for PCl₅ is 1 - (2)(0.441) = 0.117

Alternative reasonement

Assuming no dissociation we can calculate from the given data

P = nRT/V = (0.013)(0.082)(523) = 0.557 atm

Since the experimental pressure is 1 atm,  PCl₅ is dissociated. For the Daltons Law

Pt = P (PCl₅ ) + P (PCl₃ ) + P(Cl₂) = 1 atm

if x is the partial pressure due to  PCl₃ and Cl₂ we have

1 atm = P_PCl5 + 2x

We have calculated that in absence of dissociation the pressure exerted by PCl₅ should be 0.557 atm. After the dissociation the pressure for PCl₅ is:

0.557 - x (the fraction of gas dissociated)

Then

1 = 0.557 - x + 2x = 0.557 x from which x = 1 - 0.557 = 0.443 atm

0.443 is the partial pressure due to Cl₂ and PCl₃ respectively for a total of 0.886 atm and

P_PCl5 =  1 - 0.886 = 0.114

6. A thank (volume = 0.03 m³), containing nitrogen at 690 kPa, is keep opened until the pressure reach the value of 35kPa.  Assuming an isotherm process, calculate the volume of nitrogen lost into the atmosphere.

For an isotherm process P₁V₁ = P₂V₂

a) Let suppose that all the gas is expanded from 690 kPa to the atmospheric pressure (101.33 kPa)

P_iV = P_aV_a1

where P_i is the initial pressure in the thank (690 kPa), V the volume of the thank (O.03 m³) and the subscript a stand for the value of pressure and volume after the expansion. From this equation we get V_a1  = (690)(0.03)/101.33 = 0.204 m³

b) Let's suppose now that the gas left in the tank is also expanded to the atmospheric pressure

P₂V = P_aV_a2

from which V_a2  = (35)(0.03)/101.33 = 0.010 m³ The volume lost from tank is the difference between the two calculated volumes:

0.204 - 0.010 = 0.194 m³

7. Calculate the average molecular weight of air, assuming the following composition:

O₂ = 20.9%
N₂ = 78.08%

According to Avogadro law the composition in volume of a gas corresponds to its compositions in moles thus each mole of air contains 0.209 moles of O₂ and 0.7808 moles of N₂.

0.21 moles of O₂ correspond to (0.21)(32) = 6.72 g  O₂

0.78 moles of N₂ correspond to (0.7808)(28) = 21.86 g di N₂

It follows that the weight of 1 mole of air (i.e., its average molecular weight) is 6.72 + 21.86 = 28.58.

In general the average molecular weight of gas mixture can be calculated as follows

Mw_mean = x_aM_a + x_bM_b+ ...... + x_nM_n

where x is the molar fraction of a component and M the molecular weight of the component.

8. Calculate the density of air (kg/m³) at 1 atm and 70^oC. The average molecular weight of air is 29.

Thus density = (1)(29)/(0.08206)(343) = 1.03 Kg/m³

9. A package has a surface area of 3000 cm² and the packaging material has an oxygen permeability of 100 cm³/(m²)(24 h) at the 0^oC and 1 atm (STP conditions). Calculate the moles of oxygen entering the package in 24 hr.

3000  cm² correspond to 3000/(100)² = 0.3 m² the volume of O₂ entering the package in 24 h is (0.3) (100) = 30 cm³ = 0.03 liter. By using the ideal gas equation:

n = PV/RT = (1)(0.03)/(0.08206)(273) = 0.001339 moles