Respuesta :
Answer:
a) reversibly
ΔU = 0
q = 2740.16 J
w = -2740.16 J
ΔH = 0
ΔS(total) = 0
ΔS(sys) =9.13 J/K
ΔS(surr) = -9.13 J/K
b) against a constant external pressure of 1.00 atm
ΔU = 0
w = -1.66 kJ
q = 1.66 kJ
ΔH = 0
ΔS(sys) = 9.13 J/K
ΔS(surr) = -5.543 J/K
ΔS(total) = 3.587 J/K
Explanation:
Step 1: Data given:
Number of moles = 1.00 mol
Temperature = 27.00 °C = 300 Kelvin
Initial pressure = 3.00 atm
Final pressure = 1.00 atm
The gas constant = 8.31 J/mol*K
(a) reversibly
Step 2: Calculate work done
For ideal gases ΔU depends only on temperature. So as it is an isothermal (T constant).
Since the temperature remains constant:
ΔU = 0
ΔU = q + w
q = -w
w = -nRT ln (Pi/Pf)
⇒ with n = the number of moles of perfect gas = 1.00 mol
⇒ with R = the gas constant = 8.314 J/mol*K
⇒ with T = the temperature = 300 Kelvin
⇒ with Pi = the initial pressure = 3.00 atm
⇒ with Pf = the final pressure = 1.00 atm
w =- 1*8.314 *300 * ln(3)
w = -2740.16 J
q = -w
q = 2740.16 J
Step 3: Calculate change in enthalpy
Since there is no change in energy, ΔH = 0
Step 4: Calculate ΔS
for an isothermal process
ΔS (total) = ΔS(sys) + ΔS(surr)
ΔS(sys) = -ΔS(surr)
ΔS(sys) = n*R*ln(pi/pf)
ΔS(sys) = 1.00 * 8.314 * ln(3)
ΔS(sys) = 9.13 J/K
ΔS(surr) = -9.13 J/K
ΔS (total) = ΔS(sys) + ΔS(surr) = 0
(b) against a constant external pressure of 1.00 atm
Step 1: Calculate the work done
w = -Pext*ΔV
w = -Pext*(Vf - Vi)
⇒ with Vf = the final volume
⇒ with Vi = the initial volume
We have to calculate the final and initial volume. We do this via the ideal gas law P*V=n*R*T
V = (n*R*T)/P
Initial volume = (n*R*T)/Pi
⇒ Vi = (1*0.08206 *300)/3
⇒ Vi = 8.206 L
Final volume = (n*R*T)/Pf
⇒ Vf = (1*0.08206 *300)/1
⇒ Vf = 24.618 L
The work done w = -Pext*(Vf - Vi)
w = -1.00* ( 24.618 - 8.206)
w = -16.412 atm*L
w = -16 .412 *(101325/1atm*L) *(1kJ/1000J)
w = -1662.9 J = -1.66 kJ
Step 2: Calculate the change in internal energy
ΔU = 0
q = -w
q = 1.66 kJ
ΔH = 0 because there is no change in energy
Step 3: Calculate ΔS
ΔS(sys) = n*R*ln(3)
ΔS(sys) = 1.00 * 8.314 * ln(3)
ΔS(sys) = 9.13 J/K
ΔS(surr) = -q/T
ΔS(surr) = -1662.9J/300K
ΔS(surr) = -5.543 J/K
ΔS(total) = ΔS(surr) +ΔS(sys) = -5.543 J/K + 9.13 J/K = 3.587 J/K
(a) For reversible case,
- the heat added (q) is 2741.13 J,
- the work done, w is -2741.13 J,
- the change in internal energy, ΔU is 0,
- the enthalpy change, ΔH is 0,
- the entropy change of the surrounding, [tex]\Delta S_{surr}[/tex] is -9.13 J/K,
- the entropy change of the system, [tex]\Delta S_{sys}[/tex] is 9.13 J/K, and
- the total entropy change, [tex]\Delta S_{total}[/tex] is 0.
(b) For external pressure case,
- the heat added (q) is 1661.73 J,
- the work done, w is -1661.73 J,
- the change in internal energy, ΔU is 0,
- the enthalpy change, ΔH is 0,
- the entropy change of the surrounding, [tex]\Delta S_{surr}[/tex] is -5.54 J/K,
- the entropy change of the system, [tex]\Delta S_{sys}[/tex] is 9.13 J/K , and
- the total entropy change, [tex]\Delta S_{total}[/tex] is 3.6 J/K.
The given parameters:
- Number of moles, n = 1 mol
- Temperature, T = 27 °C = 27 + 273 = 300 K
- Initial pressure, P₁ = 3 atm
- Final pressure P₂ = 1 atm
- Ideal gas constant, R = 8.314 J/mol.K
(a) The reversible case:
The first law of thermodynamic is given as;
[tex]\Delta U = q + w[/tex]
where;
- ΔU is change in internal energy
For Isothermal process, ΔU = 0
q = - w
The work done on the gas is calculated as;
[tex]w = - nRT \ ln(\frac{P_i}{P_f} )\\\\w = - (1)(8.314)(300)\ ln(\frac{3}{1} )\\\\w = -2741.13 \ J[/tex]
The heat added to the system;
q = -w
q = 2741.13 J
The change in enthalpy is calculated as;
[tex]\Delta H = 0[/tex]
The change in entropy is calculated as;
[tex]\Delta S_{sys} = - \Delta S _{surr}\\\\\Delta S_{sys} = nR \ ln(\frac{P_i}{P_f} )\\\\\Delta S_{sys} = (1)(8.314)\ ln(\frac{3}{1} )\\\\\Delta S_{sys} = 9.13 \ J/K\\\\ \Delta S _{surr} = -9.13 \ J/K[/tex]
[tex]\Delta S _{total}= \Delta S _{sys}+ \Delta S _{surr}\\\\\Delta S _{total}= 9.13 - 9.13\\\\\Delta S _{total}= 0[/tex]
(b) For a constant external pressure:
[tex]w = -P \Delta V\\\\w = - P (V_f - V_i)[/tex]
The initial volume of the gas is calculated as;
[tex]P_iV_i = nRT\\\\V_i = \frac{nRT}{P_i} \\\\V_i = \frac{1 \times 8.314 \times 300}{3 atm\times (101325 \ Pa/atm)} \\\\V_i = 0.0082 \ m^3[/tex]
The final volume of the gas is calculated as;
[tex]V_f = \frac{1 \times 8.314 \times 300}{1 atm\times (101325 \ Pa/atm)} \\\\V_f = 0.0246 \ m^3[/tex]
The work done is calculated as;
[tex]w =- P(V_f - V_i)\\\\w = -101325 (0.0246 - 0.0082)\\\\w = -1661.73 \ J[/tex]
The change in internal energy;
[tex]\Delta U = 0[/tex]
The heat added to the system;
q = - w
q = 1661.73 J
The total entropy change is calculated as;
[tex]\Delta S_{surr} = \frac{-q}{T} \\\\\Delta S_{surr} =\frac{-1661.73}{300} \\\\\Delta S_{surr} =-5.54 \ J[/tex]
[tex]\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} \\\\\Delta S_{total} = 9.13 - 5.54\\\\\Delta S_{total} = 3.6 \ J/K[/tex]
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