What is the change, in joules, of the internal energy of the gas during the during the isochoric heating process, DA?
The correct answer and explanation is:
To determine the change in internal energy of a gas during an isochoric heating process (constant volume), we must understand how internal energy behaves under these conditions.
🔑 Key Concept
For an ideal gas, change in internal energy (ΔU) depends only on the temperature change and is independent of volume or pressure. During isochoric (constant volume) heating, no work is done by the gas, since: W=PΔV=0(because ΔV=0)W = P \Delta V = 0 \quad \text{(because } \Delta V = 0)
From the First Law of Thermodynamics: ΔU=Q−W\Delta U = Q – W
Since W=0W = 0, this simplifies to: ΔU=Q\Delta U = Q
So, all the heat added to the gas increases its internal energy.
✅ Formula to Use
For nn moles of an ideal monoatomic gas: ΔU=nCVΔT\Delta U = n C_V \Delta T
Where:
- CV=32RC_V = \frac{3}{2}R (for monoatomic ideal gases)
- R=8.314 J/mol\cdotpKR = 8.314 \, \text{J/mol·K}
- ΔT=Tfinal−Tinitial\Delta T = T_{\text{final}} – T_{\text{initial}}
If values for nn, TinitialT_{\text{initial}}, and TfinalT_{\text{final}} are given for process DA, you can substitute directly.
🔍 Example Calculation
Suppose:
- n=2 moln = 2 \, \text{mol}
- TD=200 KT_D = 200\, \text{K}
- TA=500 KT_A = 500\, \text{K}
Then: ΔU=nCVΔT=2⋅32⋅8.314⋅(500−200)\Delta U = n C_V \Delta T = 2 \cdot \frac{3}{2} \cdot 8.314 \cdot (500 – 200) ΔU=3⋅8.314⋅300=7482.6 J\Delta U = 3 \cdot 8.314 \cdot 300 = 7482.6 \, \text{J}
🧠 Summary
During the isochoric heating process (DA):
- No work is done.
- All added heat increases internal energy.
- Use the formula ΔU=nCVΔT\Delta U = n C_V \Delta T.
- The answer will be in joules, and depends on temperature change and number of moles.
If you have a specific diagram or values for the points D and A (temperature, pressure, or energy), I can compute the exact answer. Please upload or describe the data if available.