Question Video: Calculating the Pressure of a Gas in a New Container | Nagwa Question Video: Calculating the Pressure of a Gas in a New Container | Nagwa

# Question Video: Calculating the Pressure of a Gas in a New Container Physics

A gas in a container has a pressure of 300 kPa, a temperature of 350 K, and a volume of 0.4 m³. The gas is moved into a new container, which has a volume of 0.8 m³. No gas is lost in the process. If the temperature of the gas in the new container is 330 K, what is the pressure of the gas in the new container? Give your answer to the nearest kilopascal.

04:09

### Video Transcript

A gas in a container has a pressure of 300 kilopascals, a temperature of 350 kelvin, and a volume of 0.4 cubic meters. The gas is moved into a new container, which has a volume of 0.8 cubic meters. No gas is lost in the process. If the temperature of the gas in the new container is 330 kelvin, what is the pressure of the gas in the new container? Give your answer to the nearest kilopascal.

In this example then, we have a gas starting out inside a container that we’ll call container one. The gas is then transferred without any loss into a second larger container we’ll call container two. We want to solve for the pressure of the gas in the new container.

To begin doing that, let’s recall the ideal gas law written in terms of the number of moles of the gas. This law says that the pressure of an ideal gas multiplied by its volume is equal to the number of moles of that gas times what’s called the molar gas constant all multiplied by the gas’s temperature. The fact that in this case our gas transitions from one container with a certain set of conditions to another container with a different set means we can make two separate applications of this ideal gas law.

Our first application of this law describes the gas when it’s in container one. During this time, the gas has a pressure 𝑃 one, a volume 𝑉 one, and a temperature 𝑇 one, and there are 𝑛 one moles of the gas. Notice that we don’t have a subscript for 𝑅. This is because 𝑅 is a constant. In fact, it’s called the molar gas constant. Regardless of the conditions of our ideal gas, 𝑅 is always the same. So that’s our equation when the gas is in container one.

But now let’s consider what happens when it moves to container two. After this change, the gas will have a pressure 𝑃 two, a volume 𝑉 two, and a temperature 𝑇 two. In addition, we’ve written the number of moles of gas in the second container as 𝑛 two. We can note though that our problem statement tells us that as gas is transitioned from container one to container two, no gas is lost in that process. Since 𝑛 one and 𝑛 two refer to the number of moles of the gas in these respective cases, we know it must be the case that 𝑛 one is equal to 𝑛 two. Let’s call this number of moles, whatever it is, 𝑛. And we’ll use 𝑛 in both applications of the ideal gas law.

We want to solve for the pressure of the gas in the new container; that is, we want to solve for 𝑃 two. If we divide both sides of the second equation by 𝑉 two so that that factor cancels on the left, we get an expression where the pressure 𝑃 two is the subject. In our problem statement, we’re told the temperature of the gas in the new container, and we’re also told the volume of the new container. 𝑅, the molar gas constant, is a constant whose value we can recall. But even knowing all this, we still won’t be able to solve for 𝑃 two because we don’t know 𝑛, the number of moles of the gas.

However, notice that the same number of moles also appears in our first equation. If we take this equation and divide both sides by the molar gas constant 𝑅 times the temperature 𝑇 one, then both of those factors cancel on the right-hand side. And we now have an expression for the number of moles 𝑛. If we substitute this expression in for 𝑛 in our second equation, we find this equation for the pressure 𝑃 two. Notice that in the numerator, we’re multiplying and dividing by the molar gas constant 𝑅. So this constant cancels out. Algebraically, this expression here is the same as this slightly rearranged version. The pressure 𝑃 two is equal to the pressure 𝑃 one times the ratio of the volumes 𝑉 one over 𝑉 two times the ratio of the temperatures 𝑇 two over 𝑇 one.

In our problem statement, we’re given all of the values of the variables on the right-hand side. The pressure 𝑃 one is 300 kilopascals, the volume 𝑉 one is 0.4 cubic meters, the volume 𝑉 two is 0.8 cubic meters, the temperature 𝑇 one of the gas is 350 kelvin, and 𝑇 two is 330 kelvin. Notice that the units of cubic meters cancel from numerator and denominator, as do the units of kelvin. Our final answer will be in units of kilopascals. Calculating this expression and rounding to the nearest kilopascal, we get 141 kilopascals. This is the pressure of the gas in the new container.

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