Question Video: Calculating the Ratio of the Number of Moles of an Ideal Gas in Two Containers | Nagwa Question Video: Calculating the Ratio of the Number of Moles of an Ideal Gas in Two Containers | Nagwa

# Question Video: Calculating the Ratio of the Number of Moles of an Ideal Gas in Two Containers Physics

Two cannisters with the same volume contain the same gas at the same pressure. In cannister 1, the temperature, πβ, of the gas is 350 K. In cannister 2, the temperature, πβ, of the gas is 400 K. What is the ratio of the number of moles of the gas, πβ, in cannister 2 to the number of moles of the gas, πβ, in cannister 1?

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### Video Transcript

Two cannisters with the same volume contain the same gas at the same pressure. In cannister 1, the temperature π one of the gas is 350 kelvin. In cannister 2, the temperature π two of the gas is 400 kelvin. What is the ratio of the number of moles of the gas, π two, in cannister 2 to the number of moles of the gas, π one, in cannister 1?

Letβs say that these are our two cannisters. And weβre told that they have the same volume and contain the same gas. The number of moles of the gas in cannister 2 is π two, and the number of moles of the gas in cannister 1 is π one. We want to solve for the ratio π two to π one.

To help us do this, weβre going to treat the gas in our two cannisters as an ideal gas. This means we can describe the gas using the ideal gas law, written in terms of the number of moles of gas. This law tells us that the pressure of an ideal gas multiplied by its volume equals the number of moles of the gas times a constant multiplied by the gas temperature. We can write one expression of the ideal gas law for the gas in cannister 1 and another separate expression for the gas in cannister 2.

For the gas in cannister 1, the pressure π one times the volume π one equals the number of moles of the gas π one times the gas constant π multiplied by the gas temperature π one. π, called the molar gas constant, is indeed a constant. Thatβs why it doesnβt have a subscript in this equation. It always has the same value regardless of the ideal gas being considered. So this is our equation for the gas in cannister 1.

And now we can write a similar expression for the gas in cannister 2. We can write that the pressure of this gas π two times its volume π two equals the number of moles of the gas in cannister 2, π two, times π times the temperature π two. Since we want to solve for the ratio π two divided by π one, we can work to make π one the subject of this equation and π two the subject of this one.

Starting with our first equation, we divide both sides by the molar gas constant π multiplied by the temperature π one. Doing so means that these two factors both cancel out on the right-hand side. We find that π one is equal to π one times π one divided by π times π one. Letβs now do something similar with our second equation. In order to isolate π two, weβll divide both sides by π times π two. This causes those factors to cancel out on the right. And we find that π two equals π two times π two divided by π times π two.

Clearing some space to work on screen, letβs now take these expressions for π one and π two and substitute them in to this ratio. Doing so, we get this fraction. And if we multiply both the numerator and denominator of this fraction by π times π one, notice that in the denominator, both of those factors will cancel out. In the numerator, on the other hand, the molar gas constant π will cancel, but the temperatures remain. We get then this expression. And if we now multiply both numerator and denominator by one divided by π one times π one, then in the denominator, both π one and π one cancel out. Our fraction simplifies to π two times π two times π one divided by π one times π one times π two. Note that we can write this fraction as essentially separate ratios of pressure, volume, and temperature in our two cannisters.

Letβs now recall that our two cannisters have the same volume and that they contain gas at the same pressure. This means that π one, the volume of cannister 1, equals the volume of cannister 2, π two. And therefore, π two divided by π one equals simply one. Likewise, the pressure π one equals the pressure π two. And therefore, π two divided by π one also is one. Our ratio of moles then really comes down to the ratio of temperatures in the cannisters.

Weβre told that the temperature π one of the gas in cannister 1 is 350 kelvin, while the temperature of the gas in cannister 2 is 400 kelvin. Note that in this ratio, the units of kelvin will cancel in numerator and denominator. Written as a decimal, 350 divided by 400 is 0.875. This is the ratio of the number of moles of the gas, π two, in cannister 2 to the number of moles of the gas, π one, in cannister 1.

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