Focus on the first law of thermodynamics.
Basic example: coffee cooling down from hot to cold.
No mass transfer in or out.
Energy transfer (heat) is possible.
No mass or energy transfer.
Applies to closed systems.
Equation: Q = ΔU + W
Q: Heat transferred (joules)
ΔU: Change in internal energy (joules)
W: Work done (joules)
Significance: Describes how internal energy relates to energy transfer.
Work Equation: W = P * ΔV
Work done (joules).
Pressure (pascals).
ΔV: Change in volume (cubic meters).
Area under the PV curve represents work.
Work direction impacts sign:
Positive if volume expands (right arrow).
Negative if volume contracts (left arrow).
Heat transferred to surroundings: Q is positive.
Internal energy change:
Positive ΔU: Internal energy increases.
Work done by a system: Positive work.
Work done on a system: Negative work.
Internal energy equation: ΔU = 3 * n * R * ΔT or ΔU = 3/2 * N * kB * ΔT.
Need to know units:
ΔU: joules
ΔT: kelvin
n: number of moles
R: gas constant
N: number of molecules
kB: Boltzmann's constant
Gas undergoes isovolumetric change (volume stays constant).
Initial (A) and final (B) states known: pressure, volume, and temperature.
Find thermal energy removed when transitioning from A to B.
Understand that thermal energy means finding Q.
From the first law: Q = ΔU + W
Work done is zero in isovolumetric process.
Therefore, Q = ΔU.
Calculate ΔU using: ΔU = 3 * nR * ΔT.
Utilize the ideal gas law to determine n:
PV = nRT implies n = PV/T.
Substitute known values from state A into ΔU, taking care of temperature change (ΔT).
Example calculation yields: Q = -332,353 joules (or -332 joules for significant figures).
Key takeaways include understanding the first law of thermodynamics, conditions of closed vs. isolated systems, work definition, Clausius conventions, and applying these to problems involving internal energy and heat transfer.