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ZanzabumX [31]
3 years ago
14

Consider a mixture of air and gasoline vapor in a cylinder with a piston. The original volume is 40. cm3. If the combustion of t

his mixture releases 912 J of energy, to what volume will the gases expand against a constant pressure of 635 torr if all the energy of combustion is converted into work to push back the piston?
Chemistry
1 answer:
maks197457 [2]3 years ago
8 0

Answer:

The gases will expand until a volume of Vf= 10812 cm^3

Explanation:

Since the gas mixture is expanding at constant pressure

W=\int\limits^{Vf}_{Vi} {P} \, dV =P(Vf-Vi)

and therefore

Vf=W/P+Vi

knowing that 1 torr=133,32 Pa

p=635 torr * 133,322 Pa/torr * 1m^{3}/10^{6}cm^{3}=0,08466 J/ cm^{3}

therefore

Vf=912 J/(0,08466 J/cm^{3}) +40cm^{3} =10812 cm^{3}

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a) \Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]

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\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)

\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]

\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]

Expression of \Delta G_{max}

\frac{d \Delta G_{mixing}}{dx}=0

\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]

0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)

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\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]

\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]

\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]

\Delta G_{max}=-2721.9 J/mol

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