Respuesta :
Answer:
a. 478.69 K
b. 939.43 [tex]cm^{3}[/tex]
c. 19.30 J
d. 64.5J
Explanation:
From the question, we can identify the following;
[tex]V_{o}[/tex] = 785[tex]cm^{3}[/tex] = 0.000785 [tex]m^{3}[/tex]
[tex]T_{o}[/tex] = 400K
[tex]P_{o}[/tex] = 125 Kpa = 125 000 Pa
Using the ideal gas equation,
PV = nRT
where R is the molar gas constant = 8.314 [tex]m^{3}[/tex]⋅Pa⋅[tex]K^{-1}[/tex]⋅[tex]mol^{-1}[/tex]
Thus, n = PV/RT = (125000 × 0.000785)/(8.314 × 400) = 0.03 mol
a. Steam temperature in K
To calculate this, we use the constant pressure process;
q = nΔH
Where q is 83.8J according to the question
Thus;
83.8 = 0.03 × [34980 + 35.5[tex]T_{1}[/tex] - (34980 + 35.5[tex]T_{o}[/tex])]
83.8 = (0.03 × 35.5) ([tex]T_{1}[/tex] - 400K)
83.8 = 1.065 ([tex]T_{1}[/tex] - 400)
78.69 = ([tex]T_{1}[/tex] - 400)
[tex]T_{1}[/tex] = 400 + 78.69
[tex]T_{1}[/tex] = 478.69 K
b. Final cylinder volume
To calculate this, we make use of the Charles' law(Temperature and pressure are directly proportional)
[tex]V_{1}[/tex]/[tex]T_{1}[/tex] = [tex]V_{o}[/tex]/[tex]T_{o}[/tex]
[tex]V_{1}[/tex] = [tex]V_{o}[/tex][tex]T_{1}[/tex]/[tex]T_{o}[/tex]
[tex]V_{1}[/tex] = (785 × 478.69)/400
[tex]V_{1}[/tex] = 939.43 [tex]cm^{3}[/tex]
c. Work done by the system
Mathematically, the work done by the system is calculated as follows;
w = P([tex]V_{1}[/tex]- [tex]V_{o}[/tex]) = 125 KPa ( 939.43 - 785) = 19.30 J
d. Change in internal energy of the steam in J
ΔU = q - w = 83.8 - 19.3 = 64.5J
a. The steam temperature in Kelvin is equal to 478.69 Kelvin.
b. The final volume of the steam is equal to [tex]9.39 \times 10^{-4}\;m^3[/tex]
c. The work done by the steam is equal to 19.25 Joules.
d. The change in internal energy of the steam in Joules is 64.55 Joules.
Given the following data:
- Initial volume = 785 [tex]cm^3[/tex]
- Initial temperature = 400 K
- Initial pressure = 125 kPa
- Quantity of energy = 83.8 Joules
Scientific data:
- Ideal gas constant, R = 8.314 J/molK
a. To determine the steam temperature in Kelvin:
First of all, we would calculate the number of moles by using the ideal gas law equation:
[tex]n=\frac{PV}{RT}[/tex]
Where;
- P is the pressure.
- n is the number of moles of a gas.
- R is the ideal gas constant.
- T is the temperature.
- V is the volume.
Substituting the given parameters into the formula, we have;
[tex]n=\frac{125 \times 10^3 \times 7.85 \times 10^{-4}}{8.314 \times 400}\\\\n=\frac{98.125}{3325.6}[/tex]
n = 0.030 moles.
From the equation of heat flow:
[tex]q =n \Delta H\\\\83.8=0.030 \times [(34980+35.5T_2) - (34980+35.5T_1)]\\\\83.8=0.030 \times [34980+35.5T_2 - 34980-35.5 \times 400]\\\\83.8=0.030 \times[35.5T_2-14200]\\\\83.8=1.065T_2-426\\\\1.065T_2=426+83.8\\\\1.065T_2=509.8\\\\T_2=\frac{509.8}{1.065} \\\\T_2=478.69\;K[/tex]
b. To determine the final volume, we would apply Charles's law:
[tex]\frac{V}{T} =k\\\\\frac{V_1}{T_1} = \frac{V_2}{T_2}\\\\\frac{7.5\times 10^{-4}}{400} = \frac{V_2}{478.69} \\\\7.5\times 10^{-4} \times 478.69 = 400V_2\\\\0.3590 = 400V_2\\\\V_2 = \frac{0.3590}{400}\\\\V_2 = 9.39\times 10^{-4}\;m^3[/tex]
c. To determine the work done by the steam:
[tex]W=P \Delta V\\\\W=P(V_2-V_1)\\\\W=125 \times 10^3(9.39 \times 10^{-4}-7.85 \times 10^{-4})\\\\W=125 \times 10^3(1.54 \times 10^{-4})\\\\W=19.25\;J[/tex]
Work done = 19.25 Joules.
d. To determine the change in internal energy of the steam in Joules:
[tex]\Delta U = q- W \\\\\Delta U = 83.8-19.25\\\\\Delta U = 64.55\;Joules[/tex]
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