Answer:
2234.63
Explanation:
Work done per cycle by the engine is calculated as;
[tex]W=\int p \ dv[/tex]
#Since volume doesn't change in the isochoric steps, there is no work done, hence:
[tex]W_2=W_3=0[/tex]
#For an isothermal change of state(ideal gas):
[tex]pV=nRT\\\\W=P_iV_ i\int \frac{1}{V} \ dV\\\\=P_iV_ iIn(V_f/V_i)[/tex]
#for the expansion process:
[tex]W_i=0.5mol\times8.3145J/molK\times 700K \ In(6000cm^3/1000cm^3)\\\\W_i=5214.15J[/tex]
[tex]W_4=0.5mol\times8.3145J/molK\times 400K \ In(6000cm^3/1000cm^3)\\\\W_4=-2979.52[/tex]
[tex]W=W_1+W_2+W_3+W_4\\\\=5214.15J-2979.52\\\\=2234.63J[/tex]
Hence, the engine does 2234.63J per second.
The work done by the engine would be as follows:
[tex]2234.63[/tex] J
Given that,
Temperature [tex]= 700[/tex] K
The expansion of the cycle [tex]= 6000 cm^3[/tex]
To find,
Work Done [tex]= \int\limits[/tex][tex]p[/tex] [tex]dv[/tex]
As we know,
For Isochoric steps, the volume remains the same and hence, the
work done = 0
In order to denote the work done for the process of expansion, by putting in the values,
[tex]W_{i} = 0.5 moles[/tex] × [tex]8.3145[/tex] × [tex]700 K[/tex] × [tex]In(6000/1000)[/tex]
∵ [tex]5214.15 J[/tex] ...(i)
by putting these values at 400 K, we get,
[tex]W_{4} = -2979.52[/tex] ...(ii)
by adding (i) and (ii) as the other remain 0,
[tex]W = 2234.63 J[/tex] every second.
Thus, [tex]2234.63 J[/tex] is the correct answer.
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