Respuesta :
Answer:
Heat transfer in the process is -19.6 kJ
Work done in the process is - 28 kJ
Explanation:
As we know that the process equation is given as
[tex]P_1V_1^{1.1} = P_2V_2^{1.1}[/tex]
now we know that
[tex]P_1 = 1 bar[/tex]
[tex]V_1 = 0.2 m^3[/tex]
[tex]P_2 = 4 bar[/tex]
now from above equation we have
[tex]1(0.2)^{1.1} = 4(V)^{1.1}[/tex]
[tex]V = 0.057 m^3[/tex]
now work done in this process is given as
[tex]W = \frac{P_1V_1 - P_2V_2}{N - 1}[/tex]
here we know N = 1.1
so we have
[tex]W = \frac{10^5 (0.2) - 4 \times 10^5(0.057)}{1.1 - 1}[/tex]
[tex]W = -2.8 \times 10^4 J[/tex]
[tex]W = -28 KJ[/tex]
Now we have
[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}[/tex]
[tex]T_2 = \frac{P_2V_2 T_1}{P_1V_1}[/tex]
[tex]T_2 = \frac{4\times 10^5 (0.057) (300)}{10^5 (0.2)}[/tex]
[tex]T_2 = 342 K[/tex]
now we have
[tex]\Delta U = \frac{f}{2} n R(\Delta T)[/tex]
[tex]\Delta U = 3(P_2 V_2 - P_1 V_1)[/tex]
[tex]\Delta U = 3(4 \times 10^5(0.057) - 10^5(0.2))[/tex]
[tex]\Delta U = 8400[/tex]
so we have
[tex]Q = \Delta U + W[/tex]
[tex]Q = -28 kJ + 8.4 kJ[/tex]
[tex]Q = -19.6 kJ[/tex]
Heat transfer during the entire process is -18.8225 kJ while the Work done is -26.8625 kJ.
What is the first law of thermodynamics?
The first law of thermodynamics states that energy can neither be created nor be destroyed it can only be transformed from one form to another. therefore,
Heat transfer = Change in internal energy + Workdone
We know the pressure and volume of the gas before the process, therefore, we can find the final volume of the gas with the help of the relationship given to us,
[tex]\rm P_1 = 1\ bar\\V_1 = 0.2\ m^3\\P_2 = 4 \ bar[/tex]
[tex]P_1V_1^{1.1} = P_2V_2^{1.1}\\\\V_2 = \sqrt[1.1]{\dfrac{P_1V_1^{1.1}}{P_2}}\\\\[/tex]
Substitute the values,
[tex]V_2 = 0.0567\ m^3[/tex]
We know that the process is a polytropic process, therefore,
[tex]\rm Workdone,\ W = \dfrac{P_1V_1-P_2V_2}{N-1}[/tex]
[tex]W = \dfrac{(0.1 \times 0.2)-4 \times 0.0567}{N-1}\\\\W = -26.8625\rm\ kJ[/tex]
In order to find the heat transfer, we will use the first law of thermodynamics, therefore, we need to calculate the change in internal energy of the system,
[tex]\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}[/tex]
Substitute the values,
[tex]T_2 = 340.2\rm\ K[/tex]
Internal energy,
[tex]\triangle U = \frac{f}{2}nR(\triangle T)[/tex]
Substitute the values,
[tex]\triangle U = 3(P_2V_2-P_1V_1)\\\\\triangle U = 8.04\rm\ kJ[/tex]
According to the first law of thermodynamics,
[tex]\triangle Q = \triangle U + W\\\\\triangle Q = 8.04 -26.8625\\\\\triangle Q = -18.8225\rm\ kJ[/tex]
Hence, Heat transfer during the entire process is -18.8225 kJ while the Work done is -26.8625 kJ.
Learn more about First law of thermodynamics:
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