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Chapter 11 Gases

11.8 The Ideal Gas Law

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Dinitrogen oxide is used as an anesthetic in dentistry.

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The relationship between the four properties (P, V, n, and T) of a gas can be written equal to a constant R.

PV = RnT

Rearranging this relationship gives the ideal gas law.

PV = nRT

Ideal Gas Law

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The universal gas constant, R, can be calculated at STP using a temperature of 273 K, a pressure of 1.00 atm, a quantity of

1 mol of a gas, and a molar volume of 22.4 L.

P V R = PV = (1.00 atm)(22.4 L)

nT (1 mol) (273K) n T = 0.0821 L • atm

mol • K

Universal Gas Constant, R

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Another value for the universal gas constant is obtained using mmHg for the pressure at STP.

What is the value of R when a pressure of

760 mmHg is placed in the R value expression?

Learning Check

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What is the value of R when a pressure of 760 mmHg (at STP) is placed in the R value expression?

R = PV = (760 mmHg) (22.4 L)

nT (1 mol) (273 K)

= 62.4 L • mmHg mol • K

Solution

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Summary of Units for the Universal Gas Constant

6

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Using the Ideal Gas Law

7

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Dinitrogen oxide (N2O), laughing gas, is used by dentists as an anesthetic. If a 20.0-L tank of laughing gas contains 2.86 mol of N2O at 23 °C, what is the pressure (mmHg) in the tank?

Learning Check

Dinitrogen oxide is used as an anesthetic in dentistry.

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STEP 1 Organize the data given for the gas.

R = 62.4 L • mmHg mol • K

V = 20.0 LT = 23°C + 273 = 296 K

n = 2.86 mol

P = ?

Solution

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STEP 2 Solve the ideal gas law for the unknown. Rearrange the ideal gas law for P.P = nRT V

STEP 3 Substitute gas data and calculate the unknown quantity.

P = (2.86 mol)(62.4 L • mmHg)(296 K) (20.0 L) (mol • K)

= 2.64 x 103 mmHg

Solution (continued)

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Learning Check

A cylinder contains 5.0 L of O2 at 20 °C and 0.85 atm. How many grams of oxygen are in the cylinder?

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STEP 1 Organize the data given for the gas. P = 0.85 atm, V = 5.0 L, T = 293 K,R = 0.0821 L • atm , n ( or g =?)

mol • K

STEP 2 Solve the ideal gas law for the unknown. Rearrange the ideal gas law for n (moles).n = PV RT

Solution

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STEP 3 Substitute gas data and calculate the unknown quantity. = (0.85 atm)(5.0 L)(mol • K) = 0.18 mol of O2

(0.0821atm • L)(293 K)

Convert the moles of gas to the grams of gas using its molar mass.

= 0. 18 mol O2 x 32.00 g O2 = 5.8 g of O2

1 mol O2

Solution (continued)

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What is the molar mass of a gas if 0.250 g of the gasoccupies 215 mL at 0.813 atm and 30.0 °C?

STEP 1 Organize the data given for the gas.

R = 0.0821 L atm/mol K

P = 0.813 atm

V = 0.215 L

n = ? mol

T = 30.0 °C + 273 = 303 K

Calculating the Molar Mass of a Gas

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STEP 2 Solve the ideal gas law for the unknown. Solve for the moles (n) of gas.

n = PV = (0.813 atm) (0.215 L) mol•K = 0.00703 mol RT (0.0821 L • atm)(303 K)

STEP 3 Substitute gas data and calculate the unknown quantity.

Set up the molar mass relationship.

Molar mass = g = 0.250 g = 35.6 g/mol mol 0.00703 mol

Calculating the Molar Mass of a Gas (continued)