# Question #5b97a

##### 1 Answer

#### Answer:

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

The first thing that you need to do here is to use the **density** of the gas, **molar mass**, let's say

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

#color(blue)(ul(color(black)(PV = nRT)))#

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 theuniversal gas constant, equal to#0.0821("atm L")/("mol K")# #T# is theabsolute temperatureof 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 **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 = m/M_M * RT#

Rearrange the equation to solve for

#M_M = overbrace(m/V)^(color(blue)("the density of the gas")) * (RT)/P_ (rho)#

#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 - P_2 = "0.78 atm"#

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

If you take

#(P_1 - P_2) * V = (n_1 - n_2) * RT#

This is equivalent to

#overbrace(n_1 - n_2)^(color(blue)("the number of moles of gas released")) = (P_1 - P_2) * V/(RT)#

You already know from equation

#m_1/M_M - m_2/M_M = (P_1 - P_2) * V/(RT)#

Rearrange to get

#overbrace(m_1 - m_2)^(color(blue)("the mass of gas released")) = (P_1 - P_2) * V/(RT) * M_M#

Finally, use equation

#m_1 - m_2 = (P_1 - P_2) * V/color(red)(cancel(color(black)(RT))) * rho * color(red)(cancel(color(black)(RT)))/P_ (rho)#

which will gey you

#m_1 - m_2 = (P_1 - P_2) * V * rho * 1/P_ (rho)#

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"))))#

#color(darkgreen)(ul(color(black)(m_1 - m_2 = (6.75 * 1/P_ (rho))color(white)(.)"g")))#

All you need now is the value of

For example, at normal pressure, i.e.

#m_1 - m_2 = "6.75 g" -># atnormal pressure