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The daily production of carbon dioxide from an 780.0 MW coal-fired power plant is estimated to be 4.1480 x 104 tons (not metric).

A proposal has been made to capture and sequester the CO2 at approximately 300 K and 140 atm. At these conditions, the specific volume of CO2 is estimated to be 0.0120 m3/kg.

What volume of CO2 would be collected in a 3.00 year period? Entry field with incorrect answer x 108 m3

Respuesta :

Answer:

volume of Co_2 = 494458644.1 m^3

Explanation:

Daily production of [tex]Co_2\ is \ 4.4180\times 10^4 tons[/tex]

we know that one tons equal to 907.18 kg

so, daily production is [tex]4.4180\times 10^4 \times 907.18 kg/day[/tex]

Mass of produced Co_2 = [tex]Daily production\times 3\times 365[/tex]

                                  [tex]=4.4180\times 10^4 \times 907.18 \times 3\times 365[/tex]

                                      [tex]= 4.180\times 10^{10} kg[/tex]

[tex]Co_2[/tex] specific volume is [tex]0.012 m^3/kg[/tex]

volume of [tex]Co_2[/tex] in 3 year [tex]= Mass\ of\ produced\ Co_2\times Co_2\ specific\ volume[/tex]                      

[tex]volume =  4.180\times 10^{10} \times 0.012[/tex]

             [tex]= 494458644.1 m^3[/tex]

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