To what temperature must a balloon, initially at 22∘C and 2.25 L, be heated in order to have a volume of 10.00 L

Question

To what temperature must a balloon, initally at 22 degree C and 2.25 L, be heated in order to have a volume of 10.00 L

Answer 1

TL;DR

Start with the ideal gas law for each of the two states:

P₁V₁=n₁RT₁

P₂V₂=n₂RT₂

Rearrange and divide the equations:

n₁/n₂ = [(P₁V₁)/(RT₁)]/[(P₂V₂)/(RT₂)]

Understand n₁=n₂ because the number of moles is constant, assume P₁=P₂, and algebraically rearrange the equation:

(V₁)/(T₁) = (V₂)/(T₂)

Don't forget to convert to absolute temperature, then substitute values and solve:

(2.25L)/(295K) = (10.00L)/(T₂)

T₂ = 1311K = 1038degC

---

A full explanation starts here:

To answer this question, we'll start by assuming the gas in the balloon is ideal. The equation describing an ideal gas is

PV=nRT

P is pressure

V is volume

n is moles

R is a constant (it basically makes sure the parameters in the equation are properly proportioned. If V goes up by 1 and everything else is constant, T should not also go up by 1; the magnitude of its change is described by some number. That number is the value of R.)

T is temperature

The most unfamiliar of these is probably the term moles. In chemistry, a mole is simply a certain quantity of something. Because atoms and molecules are so very small and there are so very many of them in any physical object, it helps to define a new unit when talking about them. That unit is moles and its value is 6.02x10²³ of whatever you're talking about. A mole of carbon atoms is 6.02x10²³ carbon atoms. A mole of water molecules is 6.02x10²³ water molecules. A mole of elephants is 6.02x10²³ elephants. If I tell you I have a mole of hydrogen in a container, picture 6.02x10²³ hydrogen atoms floating around in my container. 6.02x10²³ is a strange number and you might be asking where it came from. The short answer is it doesn't matter, you probably won't need to know. The long answer is it's as many atoms of carbon as there are in 12 g of carbon-12. In other words, a mole of C-12 is defined to be 12g of C-12, and that happens to be 6.02x10²³ atoms of C-12. Why 12g of C-12? I don't know. But what I do know is it's an important number. Accordingly, we give it a cool name: Avagadro's number.

Okay, so a mole is a quantity of something; a certain number of physical things, almost always molecules or atoms. Here's the next important concept for answering this question: when the balloon changes size, it doesn't gain or lose any mass. That is, it doesn't gain or lose molecules. This means the number of moles in the balloon before and after it changes size is the same. (Formally, this is the Law of Conservation of Mass--molecules and atoms don't just randomly appear and disappear.)

Let's go back to the equation for an ideal gas: PV=nRT.

This equation is true all the time for an ideal gas, meaning it's true of the gas in the balloon when it's at 22C and 2.25L and it's true of the gas in the balloon when it's at some unknown temperature and 10L. Let's designate these states as 1 and 2 and apply them to the ideal gas equation:

Initially, the gas in the balloon is in "state 1": P₁V₁=n₁RT₁

Later, the gas in the balloon is in "state 2": P₂V₂=n₂RT₂

I didn't give R a subscript because it's a constant, so it's the same no matter what state the balloon is in.

Our task now is to relate these two equations.

We're going to use our discussion of moles to do this. Since the gas in the balloon doesn't magically gain or lose mass (or, in other words, moles) and since no one is blowing into the balloon to make it gain volume, we know the moles of gas molecules in state 1 is the same as the moles of gas molecules in state 2. In the language of math, n₁=n₂.

Now let's take a step back from the chemistry and do some algebra. We'll start by rearranging the ideal gas equations to isolate n:

In state 1: n₁=(P₁V₁)/(RT₁)

In state 2: n₂=(P₂V₂)/(RT₂)

If we divide n₁ by n₂ then we need to divide (P₁V₁)/(RT₁) by n₂, but we know n₂=(P₂V₂)/(RT₂), so we get

n₁/n₂ = [(P₁V₁)/(RT₁)]/[(P₂V₂)/(RT₂)]

Furthermore, since n₁=n₂, n₁/n₂ = 1. Of course, the R's cancel on the right side of the equation, so now we have:

1 = [(P₁V₁)/(T₁)]/[(P₂V₂)/(T₂)]

At this point, we're going to assume the pressure in the balloon is constant. Personally, I don't like using a balloon for this example because this assumption isn't intuitive. When balloons gain volume, they gain pressure. However, if you don't make this assumption, you can't solve the problem. So for the time being we'll pretend the balloon is made of some magically elastic material that doesn't constrict the gas molecules in the balloon even though the balloon is gaining volume (and the balloon material is, presumably, stretching).

If the pressure is constant, then in math language, P₁=P₂ and the equation is now:

1 = [(V₁)/(T₁)]/[(V₂)/(T₂)]

Algebraically rearranging, we get

(V₁)/(T₁) = (V₂)/(T₂)

You may recognize this as Charles Law. This isn't as important as the fact that it came from the ideal gas law. All you need is to know the ideal gas law, understand why you can cancel some terms, and rearrange the equation with algebra.

One curious observation before we continue: temperature is in the denominator. If T is ever zero, then, that side of the equation will be undefined. Mathematically we cannot have zero, but of course the physical world has no problem with zero (deg C). We have a mathematical impossibility where there is no corresponding physical impossibility. To avoid this, we must convert all temperatures to absolute temperature. We must do this even when none of the temperatures in our particular scenario are zero. (Thanks to J.R. S. for reminding me of this subtlety.)

We are now ready to insert the values we know:

V₁ = 2.25L

T₁ = 22degC+273 = 295K

V₂ = 10.00L

(2.25L)/(295K) = (10.00L)/(T₂)

T₂ = (10.00L)(295K)/(2.25L)

T₂ = 1311K = 1038degC

In addition to the balloon being magically stretchy, let's also hope it doesn't melt at 1038degC...