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For the two given changes carried out at constant pressure, state whether work is done...
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The two given changes are carried out against a constant pressure.
1/2 N_2 (g) + 3/2 H_2 (g) ---> NH_3 (g) ---- (i)
SiO_2 (s) + 3C (s) ---> SiC (s) + 2CO (g) ---- (ii)
For the first change, ∆n(gaseous) = 1 – (1/2 +3/2) = –1
That means, upon product formation, number of moles of gaseous substances decreases. So a mechanical work (pressure-volume work, more precisely) is done on the system, by the surroundings.
For the second change, ∆n(gaseous) = 2+0 – (0 +0) = 2
That means, upon product formation, number of moles of gaseous substances increases. Therefore, a mechanical pressure-volume type of work is done by the system, on the surroundings.
The amount of such work done may be calculated by the formula W = P∆V, where P is the constant external pressure, and ∆V, the change in volume.
Posted by llltkl on May 8, 2013 at 7:28 AM (Answer #1)
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