The physical state of a gaseous substance is specified by giving its volume, pressure, temperature and chemical amount. There are two equations interrelating these four variables, namely the perfect gas equation and the van der Waals equation. 2.0 mol C2H6 is confined at 273.15K in 22.414L. What is the repulsive term (in atm) of the gas under this condition when it behaves as a van der Waals gas? R = 8.314 J/molK = 0.08206 L atm/molK a = 5.507 L2atm mol-2 b = 6.51*10-2 L mol-1

Question

BlackTom AI · Accepted Answer

First, restate the problem in my own words: We have 2.0 mol of C2H6 confined at 273.15 K in a 22.414 L volume, and we want the repulsive term in the van der Waals equation under these conditions. The van der Waals equation for n moles is (P + a(n/V)^2)(V - nb) = nRT, which can be rearranged to P = nRT/(V - nb) − a(n/V)^2. Here, the repulsive term is the first term, P_rep = nRT/(V − nb).

Step 1: Identify the given quantities and constants.
- n = 2.0 mol
- T = 273.15 K
- V = 22.414 L
- b = 6.51 × 10^−2 L mol^−1
- R = 0.08206 L atm mol^−1 K^−1

Step 2: Compute nb to adjust the volume for the excluded volume.
- nb = (2.0 mol) × (6.51 × 10^−2 L/mol) = 0.1302 L

Step 3: Compute the available volume for the molecules in the repulsive term: V − nb.
- V − nb = 22.414 L − 0.1302 L = 22.2838 L

Step 4: Compute nRT for the numerator of the repulsive term.
- RT = (0.08206 L atm mol^−1 K^−1) × (273.15 K) ≈ 22.414 L atm mol^−1
- nRT = (2.0 mol) × (22.414 L atm mol^−1) ≈ 44.828 L atm

Step 5: Evaluate the repulsive term P_rep = nRT/(V − nb).
- P_rep = 44.828 L atm / 22.2838 L ≈ 2.010 atm

Result: The repulsive term under these van der Waals conditions is approximately 2.01 atm.

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