1. An ideal gas at 7°C is in a spherical flexible container having a radius of 1

Question

1. An ideal gas at 7°C  is in a spherical flexible container having a radius of 1.00 cm . The gas is heated at constant pressure to 88°C  . Determine the radius of the spherical container after the gas is heated. [Volume of a sphers =(4/3)].

2. A student adds  4.00g of dry ice (solid CO ) to an empty balloon. What will be the volume of the balloon at STP after all the dry ice sublimes (converts to gaseous CO)?

3. Given that a sample of air is made up of nitrogen, oxygen, and argon in the mole fractions0.78 N ,0.21 O , and0.010 Ar , what is the density of air at standard temperature and pressure?

4. A mixture of 1.00 g H and 1.00 g He  is placed in a  1.00-Lcontainer at 27°C . Calculate the partial pressure of each gas and the total pressure.

5. A tank contains a mixture of  52.5 g oxygen gas and 65.1g carbon dioxide gas at 27°C   . The total pressure in the tank is 8.21atm . Calculate the partial pressures of each gas in the container.

Explanation / Answer

Answer – 1) We are given, T1 = 7oC +273 = 280 K , r = 1.00 cm

T2 = 88oC +273 = 361 K

First we need to calculate the volume from the radius , r = 1.00 cm

We know the formula for volume of spheres

V = 4/3*r3

    = 4/3*3.14*(1.0 cm)3

     = 4.19 cm3

V1 = 4.19 cm3

We know Charles law

V1/T1 = V2/T2

So, V2 = V1*T2 / T1

            = 4.19 cm3 * 361 K / 280 K

            = 5.40 cm3

So, volume of the sphere after heated is V = 5.40 cm3

So, radius

3*V/4 = r3

So, 3*5.40 cm3 / 4*3.14 = r3

so, r3 = 12.90

r = 2.34 cm

2) Given, mass of CO2 = 4.00 g

Moles of CO2 = 4.00 g / 44.0 g.mol-1

                        = 0.0909 moles

We know, at STP,

1 mole = 22.4 L

So, 0.0909 moles = ?

= 2.04 L

3) We assume, 1 moles

So moles of N2 = 0.78 moles

Moles of O2 = 0.21 moles

Moles of Ar = 0.01 moles

Mass of N2 = 0.78 moles * 28.014 g/mol

                   = 21.85 g

Mass of O2 = 0.21 moles * 32.0 g/mol

                    = 6.72 g

Mass of Ar = 0.01 moles * 39.948 g/mol

                   = 0.399 g

So total mass = 21.85 + 6.72 + 0.399

                       = 28.969 g

At STP , 1 mole = 22.4 L

So, density of air = 28.969 g /22.4 L

                            = 1.29 g/L

4) Given, mass of H2 = 1.0 g

Mass of He = 1.00 g

Volume, V = 1.0 L

Moles of H2 = 1.0 g / 2.016 g.mol-1 = 0.496 mol

Moles of He = 1.0 g / 4.0026 g.mol-1 = 0.249 moles

Total moles = 0.496 + 0.249 = 0.746 moles

T = 27+273 = 300 K

We know,

PV= nRT

P = nRT/V

= 0.746 moles * 0.0821*300 K / 1.0 L

= 18.37 atm

We know, mole fraction = mole / total moles

So, mole fraction of H2 = 0.496 / 0.746 =0.665

Mole fraction of He =0.249 / 0.746 = 0.335

We know,

Partial pressure = mole fraction * total pressure

Partial pressure of H2 = 0.665 * 18.37 atm = 12.22 atm

Partial pressure of He =0.335 moles * 18.37 atm = 6.15 atm

5) mass of O2 = 52.5 g

Mass of CO2 = 65.1 g

Total pressure = 8.21 atm

Moles of O2 = 52.5 g / 32 g.mol-1 = 1.64 moles

Moles of CO2 = 65.1 g / 44 g.mol-1 = 1.48 moles

Total moles = 1.64 +1.48 = 3.12 moles

Mole reaction of O2 = 1.64 / 3.12 = 0.525

Mole reaction of CO2 = 1.48 / 3.12 = 0.478

So, partial pressure of O2 = 0.525 * 8.21 atm = 4.32 atm

partial pressure of CO2 = 0.478 * 8.21 atm =3.89 atm

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