Respuesta :
Explanation:
(a) For an isothermal process, work done is represented as follows.
W = [tex]-nRT ln(\frac{V_{2}}{V_{1}})[/tex]
Putting the given values into the above formula as follows.
W = [tex]-nRT ln(\frac{V_{2}}{V_{1}})[/tex]
= [tex]- 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})[/tex]
= [tex]-12280.82 \times ln (0.09)[/tex]
= [tex]-12280.82 \times -2.41[/tex]
= 29596.78 J
or, = 29.596 kJ (as 1 kJ = 1000 J)
Therefore, the required work is 29.596 kJ.
(b) For an adiabatic process, work done is as follows.
W = [tex]\frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}[/tex]
= [tex]\frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}[/tex]
= [tex]\frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}[/tex]
= 49.41 kJ
Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.
(c) We know that for an isothermal process,
[tex]P_{1}V_{1} = P_{2}V_{2}[/tex]
or, [tex]P_{2} = \frac{P_{1}V_{1}}{V_{2}}[/tex]
= [tex]1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})[/tex]
= 11 atm
Hence, the required pressure is 11 atm.
(d) For adiabatic process,
[tex]P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}[/tex]
or, [tex]P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}[/tex]
= [tex]1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}[/tex]
= 28.7 atm
Therefore, required pressure is 28.7 atm.
The work for isothermal process and adiabatic process is 29.596 kJ and 49.41 kJ. The final pressure for isothermal process and adiabatic process is 11 atm and 28.7 atm respectively.
Work for isothermal processes
[tex]W = -nRT ln \times \dfrac {V2}{V_1}\\\\W = -5.05 \times 8.341 \times 92.5 \ln {V_1/11}{V_1}\\\\W = -12280.82\times ln (0.09)\\\\W = -12280.82\times -2.41\\\\W = 29596.78 J\\\\W = 29.586 kJ[/tex]
(B) The work for an adiabatic process is 49. 41 J
(C)The final pressure for isothermal processes
[tex]P_1 V_1 = P_2V_2[/tex]
[tex]\bold {P_2 = \dfrac {P_1 V_1}{V_2} }\\\\\bold {P_2 = 1atm \dfrac {V_1}{V_1/11}}\\\\\bold {P_2 = 11 atm}[/tex]
(D) The final pressure for adiabatic processes,
[tex]\bold{P_1V_1^\lambda = P_2V_2^\lambda}\\\\\bold{P_2 = P_1 (\dfrac {V_1}{V_2})}\\\\\bold{P_2 = 1 atm (\dfrac {V_1}{V_1/11})^1^.^4 }\\\\\bold{P_2 = 28.7 atm }[/tex]
Therefore, The work for isothermal process and adiabatic process is 29.596 kJ and 49.41 kJ. The final pressure for isothermal process and adiabatic process is 11 atm and 28.7 atm respectively.
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