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:
<u>Step 1</u>: 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
<u>(a) reversibly</u>
<u>Step 2:</u> 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
<u>Step 3:</u> Calculate change in enthalpy
Since there is no change in energy, ΔH = 0
<u>Step 4:</u> 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
<u>(b) against a constant external pressure of 1.00 atm</u>
<u>Step 1</u>: 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
<u>Step 2:</u> Calculate the change in internal energy
ΔU = 0
q = -w
q = 1.66 kJ
ΔH = 0 because there is no change in energy
<u>Step 3: </u>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
