WEBVTT
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Two cannisters of different sizes contain the same gas at the same pressure and the same temperature.
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Cannister one contains 5.2 moles of the gas, and cannister two contains 15.6 moles of the gas.
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What is the ratio of the volume of cannister two to the volume of cannister one?
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Weβre told about these two cannisters, cannister one and cannister two.
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We donβt yet know which cannister is bigger than the other.
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But if we call the volume of cannister two π two and the volume of cannister one π one, we want to solve for this ratio: π two divided by π one.
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Weβre told that both of these cannisters contain the same gas.
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And weβre going to assume that this gas is ideal.
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Under this assumption, we can use the ideal gas law to describe this gas.
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This law holds that the pressure of an ideal gas times its volume is equal to the number of moles of the gas multiplied by the molar gas constant times the gas temperature.
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Because we have two separate cannisters, we can make two separate applications of the ideal gas law.
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For cannister one, we can write that the gasβs pressure π one times the containerβs volume π one equals the number of moles of the gas in this container π one times π
times the temperature of the gas π one.
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The reason we havenβt written a one subscript for the gas constant π
is because this is indeed a constant.
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Its value will remain the same regardless of the properties of the gas.
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Writing a similar expression for the gas in cannister two, we know that the pressure of this gas π two times the cannisterβs volume π two is equal to the number of moles of gas in cannister two π two times the gas constant π
multiplied by the gasβs temperature in this cannister, π two.
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Weβve seen that we want to solve for the ratio π two divided by π one.
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We can isolate the values of π one and π two in these two respective equations.
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If we divide both sides of our first equation by the pressure π one, then that pressure cancels on the left.
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And we find that π one equals π one times π
times π one all divided by π one.
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Likewise, dividing our second equation by π two so that that factor cancels on the left, we find that π two equals π two times π
times π two divided by π two.
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Since we now have expressions both for π one and for π two, we can substitute them into our ratio π two divided by π one.
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Clearing some space to work with this fraction, we can multiply both numerator and denominator of this fraction by one divided by the molar gas constant π
.
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Doing so causes that constant to cancel out of this expression entirely.
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Next, we can recall some important information given to us.
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Weβre told that our two cannisters contain gas which is at the same pressure and at the same temperature.
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In other words, what weβve called π one and π two are equal, and what weβve called π one and π two are also equal.
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If we recognize this fact by replacing π one and π two with π and π one and π two with π, then we can see that in both numerator and denominator of our equation, weβre multiplying by π divided by π.
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If we choose then to multiply both numerator and denominator by the inverse π divided by π, then all of these values will effectively cancel out.
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Weβre left simply with the ratio of moles of gas, π two to π one.
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We are told that cannister one contains 5.2 moles of gas, so π one equals 5.2.
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Likewise, we know that cannister two contains 15.6 moles of gas, so π two is 15.6.
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Substituting these values in for π two and π one in our equation, we find that their ratio is equal to exactly three.
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This then is the ratio of the volume of cannister two to the volume of cannister one.
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Cannister two has three times as much volume as cannister one.