# Question 5b97a

Jul 7, 2017

Here's what I got.

#### Explanation:

I don't think that you have enough information to provide a numerical solution, but you can find the mass of the gas that was released from the container in terms of ${P}_{\rho}$, the pressure at which the density of the gas is equal to ${\text{0.78 g L}}^{- 1}$.

The first thing that you need to do here is to use the density of the gas, $\rho$, to find its molar mass, let's say ${M}_{M}$, in terms of this pressure ${P}_{\rho}$.

Your starting point 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

As you know, the number of moles of gas can be expressed as the ratio between the mass of the sample, let's say $m$, and the molar mass of the gas.

n = m/M_M" "color(red)("(*)")

For the starting sample, plug this into the ideal gas law equation to get

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

Rearrange the equation to solve for ${M}_{M}$

${M}_{M} = {\overbrace{\frac{m}{V}}}^{\textcolor{b l u e}{\text{the density of the gas}}} \cdot \frac{R T}{P} _ \left(\rho\right)$

M_M = rho * (RT)/P_ (rho)" "color(darkorange)("(*)")

Now, you can use the ideal gas law equation to find the number of moles of gas that escaped from the container.

The difference between the initial pressure of the gas, let's say ${P}_{1}$, and the final pressure of the gas, let's say ${P}_{2}$, which you know to be equal to

${P}_{1} - {P}_{2} = \text{0.78 atm}$

will get you the partial pressure exerted by the gas that was released from the container.

If you take ${n}_{1}$ to be the initial number of moles of gas present in the container and ${n}_{2}$ the final number of moles present in the container, you can say that you have

$\left({P}_{1} - {P}_{2}\right) \cdot V = \left({n}_{1} - {n}_{2}\right) \cdot R T$

This is equivalent to

${\overbrace{{n}_{1} - {n}_{2}}}^{\textcolor{b l u e}{\text{the number of moles of gas released}}} = \left({P}_{1} - {P}_{2}\right) \cdot \frac{V}{R T}$

You already know from equation $\textcolor{red}{\text{(*)}}$ that you can express the number of moles of gas in terms of the mass of the sample and the molar mass of the gas, so rewrite the above equation as

${m}_{1} / {M}_{M} - {m}_{2} / {M}_{M} = \left({P}_{1} - {P}_{2}\right) \cdot \frac{V}{R T}$

Rearrange to get

${\overbrace{{m}_{1} - {m}_{2}}}^{\textcolor{b l u e}{\text{the mass of gas released}}} = \left({P}_{1} - {P}_{2}\right) \cdot \frac{V}{R T} \cdot {M}_{M}$

Finally, use equation $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{(*)}}$ to say that

${m}_{1} - {m}_{2} = \left({P}_{1} - {P}_{2}\right) \cdot \frac{V}{\textcolor{red}{\cancel{\textcolor{b l a c k}{R T}}}} \cdot \rho \cdot \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{R T}}}}{P} _ \left(\rho\right)$

which will gey you

${m}_{1} - {m}_{2} = \left({P}_{1} - {P}_{2}\right) \cdot V \cdot \rho \cdot \frac{1}{P} _ \left(\rho\right)$

Plug in the values you have to get

m_1 - m_2 = 0.78 color(red)(cancel(color(black)("atm"))) * 11.1 color(red)(cancel(color(black)("L"))) * "0.78 g" color(red)(cancel(color(black)("L"^(-1)))) * 1/(P_ (rho) color(red)(cancel(color(black)("atm"))))#

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{m}_{1} - {m}_{2} = \left(6.75 \cdot \frac{1}{P} _ \left(\rho\right)\right) \textcolor{w h i t e}{.} \text{g}}}}$

All you need now is the value of ${P}_{\rho}$, the pressure at which the density of the gas is equal to that given by the problem.

For example, at normal pressure, i.e. ${P}_{\rho} = \text{1 atm}$, the mass of the gas that was released from the container will be equal to

${m}_{1} - {m}_{2} = \text{6.75 g} \to$ at normal pressure