[huskywolf]

Calculate  work done by a system in which  formation of 1 mol CO₂ gas at 298k and 100kpa, assuming CO₂ acts as an ideal gas.

Could you please tell me if I have done this question correctly?

w=p(dv)
 
P(dv)=(dn)RT         C(s) + O₂(g) CO₂(g)

dn=1-2=-1

(dv)=[(-1)(8.314)(298)]
(dv)=-24.77

w=(100)(-24.77)= -2477 J (Exothermic)

[huskywolf]

Is this is trick question please?
The work done is just the enthalpy of formation?

Please answer .
Thank you sir

[TIGERHULL]

It's kPa, not just Pa. Also, what are you forming CO2 from, carbon and oxygen?

[unknown_analysis]

work = -P( :delta: V)

so,

work = - :delta: nRT

work = - (-1 mol)(8.314 J/mol K)(298 K)

work = 2477.572 J



_{Reference:
Petrucci, et. al. General Chemistry Principles and Modern Applications. 8th Edition. Prentice-Hall 2002}

[huskywolf]

Actually I got 2.357 atmL or ,239 Joules for the answer to this question.

PV=nRT
V= (1)(0.0821Latm/molK)(298k)/(0.987)
V=24.78L
dV= 24.78L - 22.4L = 2.38L
w=-p(dV)
w=-(0.987atm)(2.38L) =2.357 atmL or ,239 Joules?

[rabolisk]

I believe that the answer should be 0.

[huskywolf]

I believe that the answer should be 0.

Would you like to elaborate on that?

[rabolisk]

If you assume that the volume of the solid carbon is negligible compared to the volume of the gases, then you are going from 1 mole of gas to 1 mole of gas, both assumed to behave ideally. There is no volume change, and thus, no work.

[huskywolf]

If you assume that the volume of the solid carbon is negligible compared to the volume of the gases, then you are going from 1 mole of gas to 1 mole of gas, both assumed to behave ideally. There is no volume change, and thus, no work.

That makes sense,thanks.