Use the combined gas law to complete the table (assume the

Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo Tro

Problem 57P Chapter 11

Introductory Chemistry | 5th Edition

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Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo Tro

Introductory Chemistry | 5th Edition

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Problem 57P

PROBLEM 57P

Use the combined gas law to complete the table (assume the number of moles of gas to be constant)

P1

V1

T1

P2

V2

T2

1.21 atm

1.58 L

12.2 °C

1.54 atm

______

32.3 °C

721 torr

141 mL

135 K

801 torr

152 mL

______

5.51 atm

0.879 L

22.1 °C

______

1.05 L

38.3 °C

Step-by-Step Solution:

Solution 57P  

Step 1:

Here we have to Use the combined gas law to complete the table (assume the number of moles of gas to be constant). First let’s see what is combined gas law :

When we put Boyle's law, Charles' law, and Gay-Lussac's law together, we come up with the combined gas law. It shows that :

‘The ratio between the pressure-volume product and the temperature of a system remains constant’.

This can be stated mathematically as:

                = k

Where,

P = pressure,

V = volume,

T = temperature measured in kelvins,

k = constant (with units of energy divided by temperature).

For comparing the same substance under two different sets of conditions, the law can be written as:

=

Step 2:

First, let’s consider the 1st row :

We have to find V2 :

Temperature here is given in o C, let’s convert to K :

T1 = 12.2 o C + 273 = 285.2 K

T2 = 32.3 o C + 273 = 305.3 K

Using combined gas law, let’s find V2 :

V2 =

                        

V2 = 1.15 L.

Step 3:

Now, let’s consider the 2nd row :

We have to find T2 :

Here, Pressure is given in torr , let’s convert to atm :

                1 atm = 760 torr

P1 = 721 torr.

Therefore, 721 torr in atm is :

        = 721 torr

         P1= 0.948 atm.

P2 = 801 torr.

Therefore, 801 torr in atm is :

        = 801 torr

         P2= 1.053 atm.

V1 and V2 are given in mL, let’s convert to L :

V1 = 141 mL

We know, 1 L = 1000 mL

Therefore, 141 mL in L is :

            = 141 mL

                   V1 = 0.141 L

V2 = 152mL

 152 mL in L is :

            = 152 mL

        V2 = 0.152 L

Using combined gas law, let’s find T2 :

T2 =

                

                    T2 = 161.65 K.

Step 4 of 5

Chapter 11, Problem 57P is Solved
Step 5 of 5

Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo Tro
ISBN: 9780321910295

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