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

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

Key Concepts

Combined Gas Law

The combined gas law is an equation that relates pressure, volume, and temperature of a fixed amount of gas using the formula P?V?/T? = P?V?/T?. It integrates Boyle’s Law, which deals with pressure and volume, Charles’s Law, which connects volume and temperature, and Gay-Lussac’s Law, which links pressure and temperature, into a single expression that is useful when the state of a gas changes between two conditions.

Assumption of Constant Moles

This concept refers to the understanding that the number of moles of gas remains unchanged during the process. Under this assumption, the combined gas law can be applied directly since any change in pressure, volume, or temperature does not affect the amount of gas, allowing the constant n in the ideal gas law (PV = nRT) to be omitted.

Use of Absolute Temperature

When applying the combined gas law, temperature must always be expressed in an absolute scale, such as Kelvin, to ensure that the relationships among pressure, volume, and temperature are correctly maintained. This is critical because the gas laws are derived based on absolute temperature, which guarantees that computed ratios and proportions accurately reflect physical changes.

Transcript

00:01 This question asks us to use the combined gas lot to complete this table.

00:07 So the first thing that i'm going to do is standardize my units.

00:10 So i want all my pressure to be in atmospheres, my volume to be in liters, and my temperature to be in calvin.

00:16 So i'm going to copy all these values down, but convert them to the correct units in this table down here.

00:21 So starting with the pressure column, my pressure first one is already correct, 1 .21 atmospheres.

00:28 Then i have 721 tor.

00:31 So one atmosphere equals 760 tor.

00:36 So i'm going to take that 721 and divide it by 760, and that will give me 0 .949 atmospheres.

00:44 And then my last one here, 5 .51, is already in atmospheres.

00:49 My volume needs to be in liters.

00:51 So this is already correct.

00:52 1 .58 liters.

00:54 To go from mill liters to liters, one liter has 1 ,000 mil liters.

00:59 In it so i'll just divide by a thousand so i'll get 0 .141 liters here and then this is already correct 8 .879 liters and then my temperature needs to be in celsius so to get from or sorry needs to be in kelvin so to get from celsius to calvin we'll just add 273 to the celsius number so t1 here 12 .2 plus 273 is 285 .2 kelvin.

01:29 One's already correct and then 22 .1 plus 273 is 295 .1 kelvin and then we'll do the same thing with pressure here so this is good 1 .45 atmospheres 801 toward divided by 760 gives me 1 .055 atmospheres and then here 152 mils divided by 1 ,000 is 0 .152 liters and this is already correct and then 32 .3 plus 273 is, i want this right.

02:08 This is 305 .3 kelvin, and then 38 .3 plus 273 is 311 .3 kelvin.

02:18 So now that i have all of the correct units here, we'll use the combined gas law to solve for each missing variable.

02:25 So we have one missing variable per column.

02:27 So the combined gas law is p1 times v1 over t1 equals p2 times v2 over t2 so um we're just going to plug our numbers in here so for the first column p1 is 1 .21 atmospheres v1 is 1 .58 liters t1 is 285 .2 kelvin p2 is 1 .54 atmospheres we're solving for v2 and is 305 .3 kelvin.

03:03 So if we do that math, we'll take 1 .21 times 1 .58 and divide that by 285 .2.

03:10 Then we'll take that number and multiply it by 305 .3 and then divide that by 1 .54.

03:16 We'll get a v2 of 1 .33 liters...

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