(455) HL Example with thermodynamics [IB Physics HL]

Thermodynamics Overview

• Analyzing heat transfer and internal energy relationships in thermodynamics.

First Law of Thermodynamics

• Equation: Q = ΔU + W

  • Q: heat added to the system

  • ΔU: change in internal energy

  • W: work done by the system

• Area under the graph indicates work:

  • Positive work: gas expands

  • Negative work: gas is compressed.

Ideal Gas Law

• Equation: P = V * nRT

• States relationships among pressure (P), volume (V), amount of gas (n), gas constant (R), and temperature (T).

Constant Relationship

• PV^(5/3) is constant:

  • Relation between pressures and volumes at two states (A and B):

    • P_A * V_A^(5/3) = P_B * V_B^(5/3)

Example Scenario with an Ideal Monatomic Gas

1. Processes:

   • Isothermal Expansion (A to B): temperature remains constant.

   • Isovolumetric Change (B to C): volume stays the same.

   • Adiabatic Compression (C to A): no heat exchange (Q = 0).

2. Volume Relation: Volume at B is twice that of A.

   • If V_A is known, V_B = 2 * V_A.

3. Work Done:

   • Work during isothermal expansion = 208 Joules.

Part A: Thermal Energy During Expansion

• Thermal Energy (Q):

  • Q = ΔU + W

  • Isothermal means ΔU = 0 (no temperature change).

  • Therefore, Q = W = 208 Joules.

Part B: Find Temperature at Point C

• Known values:

  • P_A, V_A, T_A are given.

  • Need to find P_C, V_C, and T_C.

• Volume at C (V_C):

  • V_C = V_B (isovolumetric process), thus V_C = 1.5 * 10^-4 m³.

• Pressure at C (P_C):

  • Using P * V^(5/3):

    • P_C = P_A * (V_A/V_C)^(5/3).

  • Calculation yields P_C = 629,961 Pascals.

• Temperature at C (T_C):

  • T_C = (P_C * V_C * T_A) / (P_A * V_A), results in T_C = 193 Kelvin.

Conclusion

• Use of thermodynamic equations allows resolution of complex problems by finding interdependence between properties.