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?

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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?

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Answer 1 · 2017-07-05T04:42:11

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$ $(P_1V_1)/T_1=(P_2V_2)/T_2$ ; $V_{2} ≅ 4 \cdot L$ $V_2~=4*L$

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 $760 \cdot m m$ $760*mm$ high.

i.e. $1 \cdot a t m \equiv 760 \cdot m m \cdot H g$ $1*atm-=760*mm*Hg$

....And thus we can use the length measurement to represent pressure...We also realize that we must use $absolute temperature$ $"Absolute temperature"$ , where $absolute Temperature \equiv^{\circ} C + 273.15 \cdot K$ $"Absolute Temperature"-=""^@C+273.15*K$ .

So we solve for $V_{2} = \frac{P_{1} V_{1}}{T_{1}} \times \frac{T_{2}}{P_{2}}$ $V_2=(P_1V_1)/T_1xxT_2/P_2$ , and clearly the RHS gives an answer with units of volume.......

And so........

$\frac{771 \cdot m m \cdot H g \times 2.44 \cdot L}{295 \cdot K} \times \frac{226.1 \cdot K}{350 \cdot m m \cdot H g}$ $(771*mm*Hgxx2.44*L)/(295*K)xx(226.1*K)/(350*mm*Hg)$ ....

$≅ 4 \cdot L$ $~=4*L$

Of course, I could have reduced each pressure measurement to units of $a t m$ $atm$ , 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 $litres$ $"litres"$ , i.e. as required for a $volume$ $"volume"$ ?

Answer 2 · 2017-07-05T04:46:07

$4.18 L$ $4.18"L"$

Explanation:

As there are a constant number of moles:
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$ $("P"_1"V"_1)/"T"_1=("P"_2"V"_2)/"T"_2$

$- 47^{\circ} C$ $-47^circC$ =226^circK#

$2.44 L = 0.00244 m^{3}$ $2.44"L"=0.00244"m"^3$

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

If $\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$ $("P"_1"V"_1)/"T"_1=("P"_2"V"_2)/"T"_2$ , then $\frac{102791.546 \cdot 0.00244}{295} = \frac{46662.8288 \cdot V_{2}}{226}$ $(102791.546*0.00244)/295=(46662.8288*"V"_2)/226$ .

$V_{2} = \frac{102791.546 \cdot 0.00244 \cdot 226}{295 \cdot 46662.8288} = 0.00411777473 m^{3} \approx 0.00418 m^{3} = 4.18 L$ $"V"_2=(102791.546*0.00244*226)/(295*46662.8288)=0.00411777473"m"^3~~0.00418"m"^3=4.18"L"$ '

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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?

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