Question #265c0

Aug 22, 2017

Here's how you can do that.

Explanation:

Your starting point here will be the ideal gas law equation, which looks like this

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{P V = n R T}}}$

Here

• $P$ is the pressure of the gas
• $V$ is the volume it occupies
• $n$ is the number of moles of gas present in the sample
• $R$ is the universal gas constant, equal to $0.0821 \left(\text{atm L")/("mol K}\right)$
• $T$ is the absolute temperature of the gas

Now, let's say that the given mass of gas is $m$. The number of moles of gas present in this given mass $m$ depends on the molar mass of the gas, let's say ${M}_{M}$.

$n = \frac{m}{M} _ M$

Plug this into the ideal gas law equation to get

$P V = \frac{m}{M} _ M \cdot R T$

Next, divide both sides of the equation by $T$ to get

$\frac{P V}{T} = \frac{m}{M} _ M \cdot R$

The molar mass of the gas, which tells you the mass of exactly $1$ mole of the gas, is constant.

$\frac{P V}{T} = {\overbrace{\frac{R}{M} _ M}}^{\textcolor{b l u e}{\text{constant}}} \cdot m$

$\frac{P V}{T} = \textcolor{b l u e}{\text{constant}} \cdot m$

This means that for a given mass $m$, i.e. if you use a sample of gas of constant mass, you can say that

$\frac{P V}{T} = \textcolor{b l u e}{\text{constant}}$

This is the combined gas law equation and it tells you that for a given mass of gas $m$, you have

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2}}}$

Here

• ${P}_{1}$, ${V}_{1}$, ${T}_{1}$ are the pressure, volume, and absolute temperature of the gas at an initial state
• ${P}_{2}$, ${V}_{2}$, ${T}_{2}$ are the pressure, volume, and absolute temperature of the gas at a final state