Question

A sample of carbon monoxide occupies 2.44 L at 295.0 K and 771 torr. Find its volume at -47 degrees Celsius and 350 torr?

Answer 1

`(P_1V_1)/T_1=(P_2V_2)/T_2`; `V_2~=4*L`

Explanation:

The key to solving this problem is to realize that a mercury column is used as a highly visual measure of pressure. One atmosphere of pressure will support a column of mercury that is `760*mm` high.

i.e. `1*atm-=760*mm*Hg`

....And thus we can use the length measurement to represent pressure...We also realize that we must use `"Absolute temperature"`, where `"Absolute Temperature"-=""^@C+273.15*K`.

So we solve for `V_2=(P_1V_1)/T_1xxT_2/P_2`, and clearly the RHS gives an answer with units of volume.......

And so........

`(771*mm*Hgxx2.44*L)/(295*K)xx(226.1*K)/(350*mm*Hg)`....

`~=4*L`

Of course, I could have reduced each pressure measurement to units of `atm`, but the increase in volume is consistent with a decrease in PRESSURE, and the moderate decrease in temperature.

Do you see how the units cancel in the expression to give an answer in `"litres"`, i.e. as required for a `"volume"`?

Answer 2

`4.18"L"`

Explanation:

As there are a constant number of moles:
`("P"_1"V"_1)/"T"_1=("P"_2"V"_2)/"T"_2`

`-47^circC`=226^circK#

`2.44"L"=0.00244"m"^3`

`771"torr"=102791.546"Pa"`
`350"torr"=46662.8288"Pa"`

If `("P"_1"V"_1)/"T"_1=("P"_2"V"_2)/"T"_2`, then `(102791.546*0.00244)/295=(46662.8288*"V"_2)/226`.

`"V"_2=(102791.546*0.00244*226)/(295*46662.8288)=0.00411777473"m"^3~~0.00418"m"^3=4.18"L"`'