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Boyle’s law: V a (at constant n and T)

Va

nT

nT

nT

P

P

P

V = constant x = R

1

P

Ideal Gas EquationThis equation is a combination of 3 simpler gas relationships:

Charles’ law: VaT(at constant n and P)

Avogadro’s law: V a n(at constant P and T)

R is the gas constant

PV = nRT

What does R=?

R =

(1 atm)(22.414L)

PV

=

nT

(1 mol)(273.15 K)

The conditions 0 0C and 1 atm are called standard temperature and pressure (STP).

Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L.

PV = nRT

R = 0.082057

1 mol HCl

V =

n = 49.8 g x

= 1.37 mol

36.45 g HCl

1.37 mol x 0.0821 x 298.15 K

V =

0.950 atm

nRT

L•atm

P

mol•K

What is the volume (in liters) occupied by 49.8 g of HCl at a temperature of 25.0°C and a pressure of 0.950 atm?

T = 25.0 + 273.15 = 298.15 K

PV = nRT

P = 0.950 atm

At STP this same mass of HCl occupied a volume of 30.6 L

V = 35.3 L

1 mol HCl

V =

n = 49.8 g x

= 1.37 mol

36.45 g HCl

1.37 mol x 0.0821 x 273.15 K

V =

1 atm

nRT

L•atm

P

mol•K

What is the volume (in liters) occupied by 49.8 g of HCl at STP?

T = 0 0C = 273.15 K

PV = nRT

P = 1 atm

In the last question, this same mass of HCl at 25°C and 0.95 atm had a volume of 35.3 L

V = 30.6 L

1 mol CO2

P =

n = 25.0 g x

= 0.568 mol

44.01 g CO2

0.568 mol x 0.0821 x 296.65 K

P =

6.00 L

nRT

L•atm

V

mol•K

What is the pressure (in atmospheres) of a gas in a 6.00 L container filled with 25.0 g of carbon dioxide at 23.5°C?

T = 23.5°C+273.15 = 296.65 K

PV = nRT

V = 6.00 L

P = 2.31 atm

1 mol N2

T =

n = 15.0 g x

= 0.535 mol

28.01 g N2

6.75 atm x 2.00 L

T =

0.535 mol x 0.0821

PV

L•atm

nR

mol•K

What is the temperature (in °C) of a gas in a 2.00 L container filled with 15.0 g of nitrogen with a pressure of 6.75 atm?

P = 6.75 atm

PV = nRT

V = 2.00 L

T = 307.35 K – 273.15 = 34.2°C

n =

3.25 atm x 1.50 L

n =

0.0821 x 288.15 K

PV

L•atm

RT

32.00 g O2

mol•K

1 mol O2

What mass of oxygen gas occupies a volume of 1.50 L at 15.0 °C and 3.25 atm of pressure?

T = 15.0°C+273.15 = 288.15 K

PV = nRT

V = 1.50 L

P = 3.25 atm

0.200 mol

= 6.41 g O2

n = 0.200 mol

Ideal Gas Problem #6 (Making additional conversions)

V = 765 mL = 0.765 L

n =

P = 333 kPa = 3.29 atm

3.29 atm x 0.765 L

n =

0.0821 x 285.45 K

PV

L•atm

RT

32.00 g O2

mol•K

1 mol O2

What mass of oxygen gas occupies a volume of 765 mL at 12.3 °C and 333 kPa of pressure?

T = 12.3°C+273.15 = 285.45 K

PV = nRT

0.107 mol

= 3.44 g O2

n = 0.107 mol

Ideal Gas Problem #7 (Making additional conversions)

V = 439 mL = 4.39x10-4 L

n =

1.25 atm x 4.39x10-4 L

n =

0.0821 x 375.05 K

PV

L•atm

RT

222 g Rn

mol•K

1 mol Rn

What mass of radioactive radon gas occupies a volume of 439 mL at 101.9 °C and 950 mmHg of pressure?

T = 101.9°C+273.15 = 375.05 K

PV = nRT

P = 950 mmHg = 1.25 atm

n = 1.782x10-5 mol

1.782x10-5 mol

= 4.00x10-3 g Rn

Before & After Calculations

In some situations, we know the amounts of all 4 variables and our task is to determine one of them under new conditions where one or more of the others are changing.

Here is how we use PV=nRT for this situation:

We can use it to represent a “before” and “after” set of conditions like this:

1 = before or initial and 2 = after or final

Ideal Gas Problem #8 (Before & After)

P1 = 1.20 atm

P2 = ?

T1 = 291 K

T2 = 358 K

=

P2

P1

T2

358 K

T1

T2

T1

291 K

= 1.20 atm x

P2 = P1 x

Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18°C is heated to 85°C at constant volume. What is the final pressure of argon in the lightbulb (in atm)?

n, and V are constant so they can be left out of the equation.

Therefore:

= 1.48 atm

Dalton’s Lawof Partial Pressures

P calculations.T = PO + PH O

2

2

Bottle being filled with oxygen gas

NaClO3

Bottle full of oxygen gas and water vapor

2 NaClO3 (s) 2 NaCl (s) + 3 O2 (g)

Pressure of Collected Gas Alone calculations.

Pcollected gas = Ptotal - PH2O

Vapor pressure of water at this temperature

Total pressure of gas in container

P calculations.A =

nART

nBRT

V

V

PB =

nB

nA

cB =

cA =

nA + nB

nA + nB

mole fraction (ci) =

ni

nT

Consider a case in which two gases, A and B, are in a container of volume V.

nA is the number of moles of A

nB is the number of moles of B

PT = PA + PB

PA = cAPT

PB = cBPT

Pi = ciPT

Gas Problem #9 calculations.

0.116

8.24 + 0.421 + 0.116

A sample of natural gas contains 8.24 moles of CH4, 0.421 moles of C2H6, and 0.116 moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)?

Pi = ciPT

PT = 1.37 atm

cpropane =

= 0.0132

Ppropane = 0.0132 x 1.37 atm

= 0.0181 atm

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