Thermodynamic Processes and Calculations

Thermodynamic Processes and Ideal Gas Behavior

  • The problem discusses an ideal gas system involving air at temperature of 298.15 K298.15 \, K.
  • Given parameters:
    • Pressure, P2=3 barP_2 = 3 \, bar
    • Temperature, T2=298.15 KT_2 = 298.15 \, K
  • Objective: Calculate work, heat, changes in internal energy, and enthalpy through two defined pathways.

Key Assumptions

  • The air behaves as an ideal gas.
  • Processes involved are reversible.

Fundamental Principles

  • Changes in internal energy (ΔU\Delta U) and enthalpy (ΔH\Delta H) are state functions; hence, they will be the same for both pathways.
  • Heat (Q) and work (W) are path-dependent (trajectory functions) and can differ between pathways.

Part A: Isobaric Followed by Isochoric Process

  • Pathway Overview: First segment is isobaric (constant pressure), second is isochoric (constant volume).

Isobaric Process

  • For isobaric processes:
    • ΔH=Q\Delta H = Q
    • Enthalpy change between states 1 and A:
    1. ΔH<em>1A=C</em>P(T<em>A−T</em>1)\Delta H<em>{1A} = C</em>P (T<em>A - T</em>1)
    2. Given:
      • Pressure at A, P<em>A=P</em>1P<em>A = P</em>1 (isobaric condition)
      • Need to calculate temperature at A, TAT_A.
Ideal Gas Law Application
  • Ideal gas law: V<em>1=RT</em>1P1V<em>1 = \frac{RT</em>1}{P_1}
  • Volume at state 2: V<em>2=RT</em>2P2V<em>2 = \frac{RT</em>2}{P_2}
  • Volume at state A is equivalent to that at state 2, V<em>A=V</em>2V<em>A = V</em>2.
  • Finding temperature at state A:
    • T<em>A=P</em>AVART<em>A = \frac{P</em>A V_A}{R}
    • Substituting values yields:
    • TA≈99.4 KT_A \approx 99.4 \, K.

Calculation of m\Delta H_{1A}

  • Using values of CPC_P and temperatures:
    • ΔH1A=−574 J/mol\Delta H_{1A} = -574 \, J/mol.

Work Done in Isobaric Process

  • Work formula for isobaric process:
    • W<em>1A=−P</em>B(V<em>A−V</em>1)W<em>{1A} = -P</em>B (V<em>A - V</em>1)
    • Substitute P<em>AP<em>A for P</em>1P</em>1 in calculation:
    • Result: W1A=1653 J/molW_{1A} = 1653 \, J/mol. (positive, as volume decreases)

Change in Internal Energy ΔU1A\Delta U_{1A}

  • Apply first law of thermodynamics:
    • ΔU1A=Q+W\Delta U_{1A} = Q + W
    • Result: ΔU1A=−4131 J/mol\Delta U_{1A} = -4131 \, J/mol.

Isochoric Process (Segment A to 2)

  • Work for isochoric process: WA2=0W_{A2} = 0 (no volume change).
  • Relationship between ΔU\Delta U and Q:
    • ΔU<em>A2=Q</em>A2\Delta U<em>{A2} = Q</em>{A2}
  • Use of heat capacity at constant volume:
    • ΔU<em>A2=C</em>V(T<em>2−T</em>A)\Delta U<em>{A2} = C</em>V (T<em>2 - T</em>A)
    • Result: ΔUA2=4131 J/mol\Delta U_{A2} = 4131 \, J/mol.

Change in Enthalpy for Segment A to 2

  • Calculate ΔHA2\Delta H_{A2}:
    • ΔHA2=ΔU+Δ(PV)\Delta H_{A2} = \Delta U + \Delta (PV)
    • PVPV at volume constant and conversions yield:
    • ΔHA2=5784 J/mol\Delta H_{A2} = 5784 \, J/mol.

Total for Pathway A

  • Total work:
    • W<em>12=W</em>1A+WA2=1653+0=1653 J/molW<em>{12} = W</em>{1A} + W_{A2} = 1653 + 0 = 1653 \, J/mol.
  • Total heat:
    • Q<em>12=Q</em>1A+QA2=−5784+4131=−6053 J/molQ<em>{12} = Q</em>{1A} + Q_{A2} = -5784 + 4131 = -6053 \, J/mol.
  • Total change in internal energy:
    • ΔU<em>12=ΔU</em>1A+ΔUA2=−4131+4131=0\Delta U<em>{12} = \Delta U</em>{1A} + \Delta U_{A2} = -4131 + 4131 = 0.
  • Total enthalpy change:
    • ΔH<em>12=ΔH</em>1A+ΔHA2=−574+5784=0\Delta H<em>{12} = \Delta H</em>{1A} + \Delta H_{A2} = -574 + 5784 = 0.

Part B: Isochoric Followed by Isobaric Process

  • Pathway Overview: First segment is isochoric, second is isobaric.

Isochoric Segment 1B

  • Given the conditions:
    • W1B=0W_{1B} = 0 (isochoric implies no work)
  • Change in internal energy:
    • ΔU<em>1B=Q</em>1B=C<em>V(T</em>B−T1)\Delta U<em>{1B} = Q</em>{1B} = C<em>V (T</em>B - T_1)
    • Result: ΔU1B=12394 J/mol\Delta U_{1B} = 12394 \, J/mol.

Calculation of ΔH1B\Delta H_{1B}

  • Using enthalpy equations:
    • ΔH<em>1B=ΔU</em>1B+V(P<em>B−P</em>1)\Delta H<em>{1B} = \Delta U</em>{1B} + V (P<em>B - P</em>1)
    • Final result: ΔH1B=117052 J/mol\Delta H_{1B} = 117052 \, J/mol.

Isobaric Segment B to 2

  • Apply heat capacity and enthalpy relationships again:
    • ΔH<em>B2=C</em>P(T<em>2−T</em>B)\Delta H<em>{B2} = C</em>P (T<em>2 - T</em>B)
    • Result: ΔHB2=−17352 J/mol\Delta H_{B2} = -17352 \, J/mol.
Work Done in Isobaric Segment
  • W<em>B2=−P</em>2(V<em>2−V</em>B)W<em>{B2} = -P</em>2 (V<em>2 - V</em>B)
  • Final work calculation leads to: WB2=4958 J/molW_{B2} = 4958 \, J/mol.

Change in Internal Energy for Segment B to 2

  • Applying first law:
    • ΔU<em>B2=Q</em>B2+WB2\Delta U<em>{B2} = Q</em>{B2} + W_{B2}
    • Result: ΔUB2=−12394 J/mol\Delta U_{B2} = -12394 \, J/mol.

Total for Pathway B

  • Total work:
    • W12=0+4958=4958 J/molW_{12} = 0 + 4958 = 4958 \, J/mol.
  • Total heat:
    • Q<em>12=Q</em>1B+QB2=12394−17352=−4958 J/molQ<em>{12} = Q</em>{1B} + Q_{B2} = 12394 - 17352 = -4958 \, J/mol.
  • Total change in internal energy:
    • ΔU12=12394−12394=0 J/mol\Delta U_{12} = 12394 - 12394 = 0 \, J/mol.
  • Total enthalpy change:
    • ΔH12=117052−17352=0 J/mol\Delta H_{12} = 117052 - 17352 = 0 \, J/mol.

Conclusion

  • Noting the state functions, ΔU\Delta U and ΔH\Delta H are constant regardless of path taken under ideal gas assumption.
  • Path A (isobaric-isochoric) consumes less energy than Path B (isochoric-isobaric).
  • Recognizing differences in Q and W based on path choice due to their non-state function nature.