P v n R T Notes Scott

Ideal Gas Law

Ideal Gas Equation

Charles’ law: V T (at constant n and P)

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

Boyle’s law: V (at constant n and T)1 P

V nT

P

V = constant x = R nT

P

nT

P R is the gas constant

PV = nRT

This equation is a combination of 3 simpler gas relationships:

The conditions 0 0C and 1 atm are called

standard temperature and pressure (STP).

PV = nRT

R = PV

nT =

(1 atm)(22.414L)

(1 mol)(273.15 K)

R = 0.082057

Experiments show that at STP, 1 mole of an ideal

gas occupies 22.414 L.

What

does

R=?

L atm

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?

PV = nRT

V = nRT

P

T = 25.0 + 273.15 = 298.15 K

P = 0.950 atm

n = 49.8 g x 1 mol HCl

36.45 g HCl = 1.37 mol

V = 0.950 atm

1.37 mol x 0.0821 x 298.15 K L•atm

mol•K

V = 35.3 L

Ideal Gas Problem #1

At STP this same mass of HCl

occupied a volume of 30.6 L

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

PV = nRT

V = nRT

P

T = 0 0C = 273.15 K

P = 1 atm

n = 49.8 g x 1 mol HCl

36.45 g HCl = 1.37 mol

V = 1 atm

1.37 mol x 0.0821 x 273.15 K L•atm

mol•K

V = 30.6 L


In the last question, this same mass of HCl at

25°C and 0.95 atm had a volume of 35.3 L

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?

PV = nRT

P = nRT

V

T = 23.5°C+273.15 = 296.65 K

V = 6.00 L

n = 25.0 g x 1 mol CO2

44.01 g CO2 = 0.568 mol

P = 6.00 L

0.568 mol x 0.0821 x 296.65 K L•atm

mol•K

P = 2.31 atm


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?

PV = nRT

T = PV nR

P = 6.75 atm

V = 2.00 L

n = 15.0 g x 1 mol N2

28.01 g N2 = 0.535 mol

T = 0.535 mol x 0.0821

6.75 atm x 2.00 L

L•atm

mol•K

T = 307.35 K – 273.15 = 34.2°C


n = 0.0821 x 288.15 K

3.25 atm x 1.50 L

L•atm

mol•K

What mass of oxygen gas occupies a volume of 1.50 L at 15.0 °C

and 3.25 atm of pressure?

PV = nRT

n = PV RT

T = 15.0°C+273.15 = 288.15 K

V = 1.50 L

P = 3.25 atm

1 mol O2

32.00 g O2 = 6.41 g O2 n = 0.200 mol


0.200 mol

n = 0.0821 x 285.45 K

3.29 atm x 0.765 L

L•atm

mol•K

What mass of oxygen gas occupies a volume of 765 mL

at 12.3 °C and 333 kPa of pressure?

PV = nRT

n = PV RT

T = 12.3°C+273.15 = 285.45 K

1 mol O2

32.00 g O2 = 3.44 g O2 n = 0.107 mol

Ideal Gas Problem #6 (Making additional conversions)

0.107 mol

P = 333 kPa = 3.29 atm 1 atm

101.3 kPa

V = 765 mL = 0.765 L 1 L

1000 mL

n = 0.0821 x 375.05 K

1.25 atm x 4.39x10-4 L

L•atm

mol•K

What mass of radioactive radon gas occupies a volume of 439 L

at 101.9 °C and 950 mmHg of pressure?

PV = nRT

n = PV RT

T = 101.9°C+273.15 = 375.05 K

1 mol Rn

222 g Rn = 4.00x10-3 g Rn

n = 1.782x10-5 mol

Ideal Gas Problem #7 (Making additional conversions)

1.782x10-5 mol

P = 950 mmHg = 1.25 atm 1 atm760 mmHg

V = 439 L = 4.39x10-4 L 61 L

1 10 L

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:

Since PV = constantnT

We can use it to represent a “before” and “after”

set of conditions like this:

1 1 2 2

1 1 2 2

P V P V =

n T n T

1 = before or initial and 2 = after or final

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: P1

T1

P2

T2 =

P1 = 1.20 atm

T1 = 291 K

P2 = ?

T2 = 358 K

P2 = P1 x T2

T1 = 1.20 atm x 358 K

291 K = 1.48 atm

Ideal Gas Problem #8 (Before & After)

1 1 2 2

1 1 2 2

P V P V =

n T n T

Dalton’s Law

of Partial

Pressures

Mixtures of gases require special

handling when doing calculations.

Collected gas Evaporated

water

Ptotal = P + P Collected gas Evaporated

water

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

Bottle full of oxygen

gas and water vapor

PT = PO + PH O 2 2

NaClO3

Bottle being filled with oxygen gas

Dalton’s Law of Partial Pressures

V and T

are constant

P1 P2 Ptotal = P1 + P2

Pressure of Collected

Gas Alone

Pcollected gas = Ptotal - PH2O

Total

pressure of

gas in

container

Vapor

pressure of

water at this

temperature

Consider a case in which two gases, A and B, are in a

container of volume V.

PA = nART

V

PB = nBRT

V

nA is the number of moles of A

nB is the number of moles of B

PT = PA + PB A = nA

nA + nB B = nB

nA + nB

PA = A PT PB = B PT

Pi = i PT mole fraction ( i) =

ni

nT

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 = i PT

propane =

PT = 1.37 atm

= 0.0132

Ppropane = 0.0132 x 1.37 atm = 0.0181 atm

Gas Problem #9

0.116

8.24 + 0.421 + 0.116

3 8C H

3 8C H2 6C H4CH

Chemistry in Action:

Scuba Diving and the Gas Laws

P V

Depth (ft) Pressure

(atm)

0 1

33 2

66 3

P v n R T Notes Scott

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