Internal Energy Change in Methane Gas Under Constant Volume

Question 3: Internal Energy Change Calculation

  • System Description:

    • A 5 L tank contains 2 mol of methane gas (
      \text{CH}_4 ).
    • Initial pressure (P1) = 2 atm.
    • Final pressure (P2) = 4 atm.
    • Volume of the tank (V) = 5 L.

  • Key Concept:

    • For an ideal gas, the change in internal energy (9;9; \Delta U 9;9;) is given by the formula:
    • ΔU=nCvΔT\Delta U = n C_v \Delta T
    • Where:
    • nn = number of moles of gas
    • CvC_v = molar heat capacity at constant volume (not given, but can be related to constant pressure since volume is constant)
    •  ΔT\, \Delta T = change in temperature in Kelvin

  • Important Notes:

    • The problem states that the volume is constant. Thus, we characterize the behavior of the gas under a constant volume and varying pressure scenario.
    • Therefore, we have to define the relationship of pressure, volume, and temperature.

  • Applying the Ideal Gas Law:

    • The ideal gas law states:
    • PV=nRTPV = nRT
    • Where:
    • RR = universal gas constant = 0.0821 L·atm/(mol·K)

  • Calculating Initial Temperature (T1):

    • Using initial conditions:
    • T<em>1=P</em>1VnR=(2 atm)(5 L)(2 mol)(0.0821 L⋅atm/(mol⋅K))T<em>1 = \frac{P</em>1 V}{n R} = \frac{(2 \, \text{atm}) (5 \, \text{L})}{(2 \, \text{mol}) (0.0821 \, \text{L·atm/(mol·K)})}
    • T1=100.1642=60.9 extKT_1 = \frac{10}{0.1642} = 60.9 \, ext{K} (approximately)

  • Calculating Final Temperature (T2):

    • Now calculating the final temperature with final pressure:
    • T<em>2=P</em>2VnR=(4 atm)(5 L)(2 mol)(0.0821 L⋅atm/(mol⋅K))T<em>2 = \frac{P</em>2 V}{n R} = \frac{(4 \, \text{atm}) (5 \, \text{L})}{(2 \, \text{mol}) (0.0821 \, \text{L·atm/(mol·K)})}
    • T2=200.1642=121.8 extKT_2 = \frac{20}{0.1642} = 121.8 \, ext{K} (approximately)

  • Calculating Change in Temperature (ΔT):

    • ΔT=T<em>2−T</em>1=121.8 K−60.9 K=60.9 K\Delta T = T<em>2 - T</em>1 = 121.8 \, K - 60.9 \, K = 60.9 \, K

  • Calculating Internal Energy Change (ΔU):

    • We assume C<em>vC<em>v can be related to C</em>pC</em>p by the relation of ideal gases:
    • C<em>v=C</em>p−RC<em>v = C</em>p - R
    • Cv=34 J/(mol K)−0.0821 J/(mol K)C_v = 34 \, J/(mol \, K) - 0.0821 \, J/(mol \, K) (1 atm·L = 101.325 J, so conversions needed if using this unit) => approximate values converted.
    • Therefore:
    • Cvext(approximately)=33.918 J/(mol K)C_v ext{ (approximately)} = 33.918 \, J/(mol \, K)
    • Now to calculate:
    • ΔU=nimesCvimesΔT=(2 extmol)×(33.918 J/(mol K))×(60.9 K)\Delta U = n imes C_v imes \Delta T = (2 \, ext{mol}) \times (33.918 \, J/(mol \, K)) \times (60.9 \, K)
    • ΔU=2×33.918 J/(mol K)imes60.9≈4133.06 J\Delta U = 2 \times 33.918\, J/(mol \, K) imes 60.9 \approx 4133.06 \, J

  • Final Result:

    • The total internal energy change during the process is approximately:
    • ΔU≈4133 J\Delta U \approx 4133 \, J