B) Volume:
[tex]V_a=0.0062 m^3\\V_b=0.012 m^3\\V_c=0.019 m^3\\V_d=0.0093 m^3[/tex]
C) Heat:
[tex]Q_1=+3474 J\\Q_2=0\\Q_3=-2808 J\\Q_4=0[/tex]
D) Work:
[tex]W_1=+3474J\\W_2=+1944 J\\W_3=-2808 J\\W_4=-1944 J[/tex]
E) Change in internal energy:
[tex]\Delta U_1 = 0\\\Delta U_2 = -1944 J\\\Delta U_3=0\\\Delta U_4 = +1944 J[/tex]
F) Efficiency:
[tex]e_1=19\%\\e_2=25\%[/tex]
Explanation:
Missing figure: find it in attachment.
B)
We find the volume at point A by using the equation of state for ideal gas:
[tex]P_a V_a = nRT_A[/tex]
where
[tex]P_a=8.4 atm =8.51\cdot 10^5 Pa[/tex] is the gas pressure
[tex]V_a[/tex] is the gas volume
n = 1 mol is the number of moles
[tex]R=8.31 J/mol K[/tex] is the gas constant
[tex]T_a=360^{\circ}+273=633 K[/tex] is the gas temperature
So we find
[tex]V_a=\frac{nRT_a}{p_a}=\frac{(1)(8.31)(633)}{8.51\cdot 10^5}=0.0062 m^3[/tex]
Process 1 is a isothermal process (constant temperature). So we can apply Boyle's law:
[tex]P_a V_a = P_b V_b[/tex]
where
[tex]P_a = 8.4 atm[/tex] is the initial pressure
[tex]P_b = 4.2 atm[/tex] is the final pressure
[tex]V_a=0.0062 m^3[/tex] is the initial volume
Then we find the volume in B:
[tex]V_b=\frac{p_a V_a}{p_b}=\frac{(8.4)(0.0062)}{4.2}=0.012 m^3[/tex]
The volume at C can be found by using the equation of state, as done for the volume at A:
[tex]p_c V_c = nR T_C[/tex]
with
[tex]p_c = 2.0 atm = 2.03\cdot 10^5 Pa[/tex]
[tex]T_c=200^{\circ}+273=473 K[/tex]
So the volume in C is
[tex]V_c=\frac{nRT_c}{p_c}=\frac{(1)(8.31)(473)}{2.03\cdot 10^5}=0.019 m^3[/tex]
Finally, process 3 is at a constant temperature, so we can use Boyle's law again:
[tex]p_c V_c = p_d V_d[/tex]
with
[tex]p_d=4.1 atm = 4.15\cdot 10^{5} Pa[/tex]
So the volume in D is
[tex]V_d=\frac{p_c V_c}{p_d}=\frac{(2.03\cdot 10^5)(0.019)}{4.15\cdot 10^5}=0.0093 m^3[/tex]
C)
Process 1 is at constant temperature, so the change in internal energy is zero; so keeping in mind 1st law of thermodynamics:
[tex]\Delta U = Q-W[/tex]
the heat exchanged is equal to the work done:
[tex]Q_1=W_1 = nRT_1 ln \frac{V_b}{V_a}[/tex]
where
[tex]T_1=633 K\\V_a=0.0062 m^3\\V_b=0.012 m^3[/tex]
Substituting,
[tex]Q_1=(1)(8.31)(633)ln\frac{0.012}{0.0062}=3474 J[/tex]
Process 2 is adiabatic (no heat exchanged), so [tex]Q_2=0[/tex]
Process 3 is also isothermal, so the heat exchanged is
[tex]Q_3=W_3=nRT_3 ln \frac{V_d}{V_c}[/tex]
where
[tex]T_3=473 K\\V_c=0.019 m^3\\V_d=0.0093 m^3[/tex]
So
[tex]Q_3=(1)(8.31)(473)ln\frac{0.0093}{0.019}=-2808 J[/tex]
Process 4 is adiabatic, so the heat exchanged is zero: [tex]Q_4=0[/tex]
D)
Process 1 is isothermal, so the work done is equal to the heat exchanged:
[tex]W_1=Q_1=+3474 J[/tex]
Process 2 is adiabatic, so work done is equal to the negative of the change in internal energy, therefore:
[tex]W_2=-\Delta U_2=-\frac{3}{2}nR(T_3-T_1)[/tex]
where
[tex]T_3=473 K\\T_1=633 K[/tex]
Substituting,
[tex]W_2=-\frac{3}{2}(1)(8.31)(473-633)=+1944 J[/tex]
Process 3 is isothermal, so the work done is equal to the heat exchanged:
[tex]W_3=Q_3=-2808 J[/tex]
Process 4 is adiabatic, so work done can be calculated as for process 2:
[tex]W_4=-\Delta U_4 = -\frac{3}{2}nR(T_1-T_3)[/tex]
Substituting,
[tex]W_4=-\frac{3}{2}(1)(8.31)(633-473)=-1944J[/tex]
E)
The change in internal energy can be determined using
[tex]\Delta U = Q-W[/tex]
For process 1,
[tex]\Delta U_1 = Q_1 - W_1 = +3474 -3474 = 0[/tex] (in an isothermal process, change in internal energy is zero)
For process 2,
[tex]\Delta U_2 = Q_2-W_2=0-1944 = -1944 J[/tex]
For process 3,
[tex]\Delta U_3 = Q_3-W_3=-2808-(-2808)=0[/tex] (isothermal process)
For process 4,
[tex]\Delta U_4 = Q_4-W_4=0-(-1944)=+1944 J[/tex]
F)
The efficiency of the cycle is given by
[tex]e_1=\frac{W}{Q_{in}}[/tex]
where
[tex]W=W_1+W_2+W_3+W_4=+3474+1944-2808-1944=+666 J[/tex] is the net work done
[tex]Q_{in}=Q_1=+3474 J[/tex] is the heat in input
Substituting,
[tex]e_1=\frac{666}{3474}=0.19=19\%[/tex]
The efficiency of a Carnot cycle operating within the same temperatures is
[tex]e_2=1-\frac{T_L}{T_H}[/tex]
where
[tex]T_L=473 K\\T_H=633 K[/tex]
Substituting,
[tex]e_2=1-\frac{473}{633}=0.25=25\%[/tex]
Learn more about ideal gases:
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