Question Video: Calculating the Volume of an Ideal Gas Physics

A cloud of gas has a pressure of 220 kPa and a temperature of 440 k. The gas contains 8.2 moles of a particle with a molar mass of 10.5 g/mol. Find the volume of the cloud. Use 8.31 m²⋅kg/s²⋅k⋅mol for the value of the molar gas constant. Give your answer to two decimal places.

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Video Transcript

A cloud of gas has a pressure of 220 kilopascals and a temperature of 440 kelvin. The gas contains 8.2 moles of a particle with a molar mass of 10.5 grams per mole. Find the volume of the cloud. Use 8.31 meter squared kilogram per second squared kelvin mole for the value of the molar gas constant. Give your answer to two decimal places.

Okay, so in this question we’re being asked to find the volume of a cloud of gas. To help us work this out, we’ve been given a load of other information about the cloud of gas, including its pressure, its temperature, the number of moles of gas, and the molar mass of the gas particles. Let’s label the pressure as capital 𝑃, the temperature as capital 𝑇, the number of moles as 𝑛, and the molar mass as capital 𝑀. We are also given a value to use for the molar gas constant, and we give this constant the symbol capital 𝑅. The last quantity we need to give a name to is the volume of the gas cloud, which is what we want to work out. And let’s label this volume as capital 𝑉.

In order to answer this question, let’s recall a formula that’s known as the ideal gas law. This law says that the pressure 𝑃 of a gas multiplied by its volume 𝑉 is equal to 𝑛, the number of moles, multiplied by the molar gas constant 𝑅 multiplied by the gas temperature 𝑇. In this equation, we know the value of the pressure 𝑃. We also know the number of moles 𝑛 of the gas. We’ve got a value for the molar gas constant 𝑅. And we know the gas temperature. 𝑇. We can see then that there’s only one unknown quantity in this equation, and that’s the volume 𝑉 of the gas, which is what we want to work out in this question. We can therefore rearrange this equation in order to make the volume 𝑉 the subject. And then if we substitute in our values for 𝑃, 𝑛, 𝑅, and 𝑇, we’ll be able to calculate the volume of this cloud of gas.

It’s worth briefly noticing that this equation involves the pressure, the volume, the number of moles, the molar gas constant, and the temperature of the gas. However, it does not involve the molar mass of the gas. This means that we’re not actually going to need to use this value for the molar mass, capital 𝑀, in order to answer this question. To calculate the volume of a cloud of gas, we simply need to know its pressure, its temperature, and the number of moles of gas, along with our value for the molar gas constant. Now, we said that the first step is to make 𝑉 the subject of the ideal gas law equation. To do this, we divide both sides of the equation by the gas pressure 𝑃.

Then we can see that on the left-hand side of the equation, the 𝑃 in the numerator cancels with the 𝑃 in the denominator. This leaves us with an equation that says 𝑉 is equal to 𝑛 times 𝑅 times 𝑇 divided by 𝑃. Now, before we go ahead and substitute values into the right-hand side of this equation, we should take a moment to think about the units of these quantities. In the numerator of the fraction, we’ve got the units of 𝑛, which is moles, multiplied by the units of the molar gas constant 𝑅, which is meter squared kilogram per second squared kelvin mole, multiplied by the units of temperature 𝑇, which is kelvin. We can see that the moles cancel from the numerator and denominator, and so do the units of kelvin.

This leaves us with overall units for the numerator on the right-hand side that we can write as kilograms meters squared per second squared. In the denominator on the right-hand side, we just got the quantity pressure. Now we’ve been given a pressure value in units of kilopascals. But the SI unit for pressure is pascals. Since pressure is defined as force divided by area, where force has SI units of newtons and area has SI units of meter squared, then we can see that units of pascals must be equivalent to units of newtons per meter squared.

Then we can also recall Newton’s second law of motion tells us that force is equal to mass multiplied by acceleration. The SI unit of mass is the kilogram and the SI unit of acceleration is meters per second squared. This means then that units of newtons must be equal to units of kilograms meters per second squared.

If we then replace the newtons in this expression for units of pascals by this expression here, we get this expression here for units of pascals. We can write this expression more neatly like this. And then we see that we can cancel one factor of meters from the numerator and denominator. We have then that units of pascals are equal to units of kilograms per meter second squared. Overall then on the right-hand side of this equation, we have units of kilograms meters squared per second squared in the numerator. And if we measure the pressure 𝑃 in units of pascals, then the denominator has units of kilograms per meter second squared.

The units on the right-hand side of the equation must be equal to the units of the numerator divided by the units of the denominator. Now we see that we can cancel out both the units of kilograms and the units of per second squared. This leaves us with units of meter squared divided by one over meters. This can be written more simply as units of meters cubed, which we can then recognize as the SI unit for the quantity volume. The outcome of all this discussion about units then is that if we use the quantities 𝑛, 𝑅, and 𝑇 on the right-hand side of this expression with the units that they’ve been given to us in, then in order to calculate a volume 𝑉 with units of meters cubed, we’ll need a pressure 𝑃 in units of pascals.

To convert our value of pressure from units of kilopascals into units of pascals, we can recall that one kilopascal is equal to 1000 pascals. That means that to convert a value from kilopascals into pascals, we multiply it by 1000. So then in units of pascals, the pressure 𝑃 is equal to 220 multiplied by 1000 pascals. This works out as a pressure of 220000 pascals. So now that we’ve got a value for the pressure in units of pascals, we know that if we substitute this value along with our values for 𝑛, 𝑅, and 𝑇 in the units that they’re given into this equation, then we’ll calculate a volume 𝑉 with units of meters cubed. Let’s clear some space on the screen so that we can do this.

We’ve got this equation here for the volume 𝑉 of the gas cloud in terms of the quantities 𝑛, 𝑅, 𝑇, and 𝑃. Substituting our values for these quantities in their SI units, we get this expression for the volume 𝑉. In the numerator, we have 8.2, which is the value for 𝑛 in units of moles, multiplied by 8.31, which is the value for 𝑅 in units of meters squared kilogram per second squared kelvin mole, multiplied by 440, which is the value for the temperature 𝑇 in units of kelvin. Then in the denominator of the fraction, we’ve got 220000, which is the value for the pressure 𝑃 in units of pascals. Notice that in this expression we haven’t written out all of these individual units, but rather we’ve just written the units of meters cubed that we know the volume 𝑉 will have.

We can then evaluate this expression by typing it into a calculator. When we do this, we find that 𝑉 is equal 0.136284 meters cubed. However, we should take care to notice that we’re asked to give our answer to two decimal places. To do this, we look at the value in the third decimal place, which in this case is six. Since this value is greater than or equal to five, then the value in the second decimal place gets rounded up. Rounding this number then gives us our answer to the question that, to two decimal places, the volume of the gas cloud is equal to 0.14 meters cubed.

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