Question #d132b
1 Answer
Explanation:
Your starting point here will be the ideal gas law equation, which looks like this
#color(blue)(bar(ul(|color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#
where
#P# - the pressure of the gas
#V# - the volume it occupies
#n# - the number of moles of gas
#R# - the universal gas constant, usually useful as#0.0821 ("atm" * "L")/("mol" * "K")#
#T# - the absolute temperature of the gas
Now, the idea here is that you need to use the fact that the number of moles of gas can be written in terms of the mass of a given mass, let's say
You will have
#n = m/M_M" " " "color(orange)((1))#
Next, use the fact that the density of the gas is equal to the mass of the given sample, which we've said to be equal to
You will have
#rho = m/V " " " "color(orange)((2))#
Plug equation
#PV = m/M_M * RT#
This is equivalent to
#P * M_M = m/V * RT#
Now use equataion
#P * M_M = rho * RT#
This gets you
#color(purple)(bar(ul(|color(white)(a/a)color(black)(rho = P/(RT) * M_M)color(white)(a/a)|)))#
The molar mass of methane is equal to
#rho = (3.5 color(red)(cancel(color(black)("atm"))))/(0.0821 ( color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (273.15 + 127)color(red)(cancel(color(black)("K")))) * "16.04 g" color(red)(cancel(color(black)("mol"^(-1))))#
#color(green)(bar(ul(|color(white)(a/a)color(black)(rho = "1.7 g L"^(-1))color(white)(a/a)|)))#
The answer is rounded to two sig figs.