Question

Part A Background Information; Solution already found!! Calculate the change in enthalpy (ΔH) for 1.0 mol of carbon dioxide (CO2) initially at a volume of 5.00 L and a temperature of 298 K to a final volume of 10.00 L and a temperature of 373 K. Explicitly state all assumptions along with a justification for any and all assumptions made. Express your answer in units of kilojoules per mole (kJ/mol)

The part i need help with;

Part B: The enthalpy is a common form of energy used in biochemical thermodynamics and is a state function, meaning the change in enthalpy (ΔH) for the process above should be independent of the path. Explicitly illustrate this point by calculating the change in enthalpy for the expansion process above along a different path than you did in part (A) of this problem

Not sure how to calculate along a different path. please show work.

Answer 1

The change in enthalpy at constant pressure can be calculated using the following formulas

Answer 2

Solution for Part B: Illustrating Path Independence of Enthalpy Change

Objective:
Calculate the change in enthalpy (ΔH) for the expansion of CO₂ (from 5.00 L at 298 K → 10.00 L at 373 K) using a different thermodynamic path than in Part A, demonstrating that ΔH is a state function (path-independent).

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Key Assumptions (Same as Part A):

1. CO₂ behaves as an ideal gas: Valid at moderate pressures and temperatures.

2. Constant heat capacity ($C_p$): $C_p$ for CO₂ ≈ 37.1 J/mol·K (assumed independent of temperature).

3. Isenthalpic process for expansion: No work is done against external pressure in free expansion (Joule-Thomson effect negligible for ideal gases).

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Alternative Path:

Path 2 (Two-Step Process):

1. Step 1: Isothermal Expansion (298 K, 5.00 L → 10.00 L)

• For an ideal gas, ΔU = 0 in isothermal processes (no temperature change).

• Work done ($w$) = $-nRT\ln (V_2\mathrm{/}{V}_1)$:

  $$w=-(1\text{mol})(8.314$$

• Heat added ($q$) = $-w=+1718\text{J}$ (to maintain ΔU = 0).

• ΔH₁ = 0 (since ΔH = ΔU + Δ(PV), and Δ(PV) = 0 for isothermal ideal gas).

• 
  

2. Step 2: Isochoric Heating (10.00 L, 298 K → 373 K)

• At constant volume, $w=0$.

• Heat added ($q_v$) = $n{C}_v\mathrm{\Delta }T$:

  $$C_v=C_p-R=37.1-8.314=28.8$$$$q_v=(1\text{mol})(28.8$$

• ΔH₂ = nC_pΔT:

  $$\mathrm{\Delta }{H}_2=(1\text{mol})(37.1$$

3. Total ΔH for Path 2:

   $$\mathrm{\Delta }{H}_{\text{Path 2}}=\mathrm{\Delta }{H}_1+\mathrm{\Delta }{H}_2=0+2782.5\text{J}=2.78\text{kJ/mol}$$

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Comparison with Part A (Direct Path):

In Part A, ΔH was calculated directly using:

$$\mathrm{\Delta }H=n{C}_p\mathrm{\Delta }T=(1\text{mol})(37.1$$

Result:
Both paths yield identical ΔH values, proving that enthalpy is a state function.

Answer:

The change in enthalpy (ΔH) for the process via the alternative path is:

$$\boxed{{\displaystyle 2.78\text{kJ/mol}}}$$