### Video Transcript

The volume of an ideal gas is found to be 200 milliliters at STP. At 23 degrees Celsius, which of the following expressions gives the pressure of the gas if its new volume is 550 milliliters?

And we’ve been given five expressions that use some or all of these numbers in various combinations. STP is standard temperature and pressure. Since 1982, IUPAC have defined STP to mean zero degrees Celsius and one bar of pressure. One atmosphere is equivalent to 1.01325 bar. Other sources state that STP is zero degrees Celsius and one atmosphere.

So for this question, it’s important we know which definition is being used. The pressure unit in the answers is millimeters of mercury. Which is a unit of measurement derived from looking at the difference in height between two connected columns of mercury. If there’s a gas pressure difference between the two columns, one of the columns will have more mercury in it. And the difference in height of the two columns of mercury can be used to quantify that pressure difference.

A pressure difference of one atmosphere is equivalent to 760 millimeters of mercury. That’s the number we see in all five of the expressions, which suggests we’re using one atmosphere as our standard pressure.

Now let’s have a look at the other numbers in the expression. As we’ve already said, 760 derives from the initial pressure. 200 comes from the initial volume in milliliters, while 550 comes from the final volume. In the third and fourth expressions, the initial temperature, zero degrees Celsius, has been kept in its original units. But when we’re looking at gas relationships, we need to be using absolute temperature scales, like the Kelvin scale.

If we evaluated the third expression, we’d get zero because we’d be multiplying by zero. And if we evaluated the fourth expression, we’d get an invalid answer because we can’t divide by zero. Instead, we should be using Kelvin temperatures like that we see in the fifth expression, where 273 is the initial temperature in Kelvin.

To convert degrees Celsius to Kelvin, we simply add 273. So zero degrees Celsius is 273 Kelvin. And the final unknown is 296, which comes from the final temperature, which is 23 plus 273. So we now understand what all these expressions are trying to do. They’re combining the initial and final temperatures and the initial and final volumes and the initial pressure to try and produce the final pressure. But how do we find which one is correct?

To help us on our way, we can use the combined gas law, which tells us the relationship between the pressure, volume, and temperature of a gas before and after a change. The question mentions we’re dealing with an ideal gas, which is a hint that we could use the ideal gas law. So if you can’t remember the combined gas law, you can rearrange the ideal gas law.

If the amount of gas doesn’t change, then 𝑛 times 𝑅, the gas constant, is a constant. This means that even if we change the conditions for our sample, we’ll always be able to compare the pressure, volume, and temperature using this formula. So all we need to do is rearrange the equation in terms of the final pressure.

The final pressure is equal to the initial pressure multiplied by the initial volume divided by the final volume multiplied by the final temperature divided by the initial temperature. All three of the remaining expressions have the right number for the initial pressure. However, the fifth expression has the final volume and the initial volume the wrong way up. It should be 200 divided by 550. And the second expression has the final temperature and initial temperature the wrong way up.

So the first expression, if evaluated, would give us our final pressure. Because the volume increases, the pressure goes down. But because the temperature increases, the pressure goes up. But the scale of the increase in the volume is much greater than that for the temperature. So our pressure is going to be meaningfully less than 760 millimeters of mercury. So our final expression is 760 times 200 divided by 550 times 296 divided by 273 millimeters of mercury.