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 |

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.