Thermodynamics: Comprehensive NEET Physics Formulae Summary

1. Key Formulae and Explanations

1.1 Zeroth Law of Thermodynamics

• Formula:
• • There is no direct mathematical formula, but it states that if two systems are in thermal equilibrium with a third system, they are in thermal equilibrium with each other.
• Explanation:
• • This law introduces the concept of temperature as a fundamental and measurable property that dictates whether systems are in thermal equilibrium.

NEET Tip:

In NEET, the Zeroth Law can be used to understand the concept of temperature scales and thermometers.

1.2 First Law of Thermodynamics

• Formula: $\Delta Q=\Delta U+\Delta W$
• Explanation:
• • $\Delta Q$: Heat supplied to the system
  • $\Delta U$: Change in internal energy of the system
  • $\Delta W$: Work done by the system on its surroundings
  • This law is a statement of the conservation of energy, where the heat added to the system is used to increase its internal energy and perform work.

Example Application:

A gas in a piston absorbs 500 J of heat and does 200 J of work. The change in internal energy is calculated as: $\Delta U=\Delta Q-\Delta W=500J-200J=300J$

Common Mistake:

Confusing heat ($\Delta Q$) with internal energy ($\Delta U$). Remember that heat is energy in transit, not stored energy.

1.3 Specific Heat Capacity

• Formula: $s=\frac{1}{m}\frac{\Delta Q}{\Delta T}$
• Explanation:
• • $s$: Specific heat capacity (J/kg·K)
  • $m$: Mass of the substance (kg)
  • $\Delta Q$: Heat added or removed (J)
  • $\Delta T$: Change in temperature (K)
  • This formula gives the amount of heat required to raise the temperature of a unit mass of the substance by one degree Kelvin.

Real-life Application:

Water's high specific heat capacity makes it an excellent coolant in car engines.

Mnemonic:

"Specific Heat Stays Constant," to remember that specific heat capacity is a property intrinsic to the material, not the mass.

1.4 Work Done in Isothermal Process

• Formula: $W=\mu RT\ln \frac{V_2}{V_1}$
• Explanation:
• • $W$: Work done by the gas (J)
  • $\mu$: Number of moles of gas
  • $R$: Universal gas constant (8.314 J/mol·K)
  • $T$: Absolute temperature (K)
  • $V_1,V_2$: Initial and final volumes (m³)
  • This formula is used to calculate the work done by an ideal gas when it expands or compresses isothermally.

NEET Problem-Solving Strategy:

In NEET, be careful to distinguish between isothermal and adiabatic processes when calculating work done, as the formulas differ significantly.

1.5 Work Done in Adiabatic Process

• Formula: $W=\frac{\mu R(T_1-T_2)}{\gamma -1}$
• Explanation:
• • $\gamma$: Ratio of specific heats ($C_p/C_v$)
  • @@T_1, T_2@@: Initial and final temperatures (K)
  • Adiabatic processes occur without heat exchange with the surroundings. The work done leads directly to a change in internal energy.

Common Misconception:

Students often confuse isothermal and adiabatic processes. Remember, adiabatic means no heat exchange ($\Delta Q=0$).

1.6 Second Law of Thermodynamics

• Formula (Efficiency of Carnot Engine): $\eta =1-\frac{T_2}{T_1}$
• Explanation:
• • $\eta$: Efficiency
  • $T_1,T_2$: Temperatures of the hot and cold reservoirs (K)
  • This formula defines the maximum efficiency of any heat engine operating between two temperatures.

NEET Exam Strategy:

Understanding the Carnot engine's efficiency is crucial for solving questions on thermodynamics in NEET.

Quick Recap:

• Zeroth Law: Basis of temperature definition.
• First Law: Energy conservation ($\Delta Q=\Delta U+\Delta W$).
• Specific Heat Capacity: Heat required per unit mass to raise the temperature.
• Work in Isothermal Process: $W=\mu RT\ln \frac{V_2}{V_1}$
• Work in Adiabatic Process: $W=\frac{\mu R(T_1-T_2)}{\gamma -1}$
• Second Law: Maximum efficiency of heat engines.

Practice Questions:

1. Question: A gas is compressed adiabatically from a volume of 4 m³ to 1 m³. If the initial temperature was 300 K, find the final temperature assuming $\gamma =1.4$. Solution: Apply the formula $T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}$ to solve for $T_2$.
2. Question: Calculate the heat required to raise the temperature of 2 kg of water from 20°C to 80°C. $s_{water}=4186\text{J/kgcdotpK}$. Solution: Use $\Delta Q=ms\Delta T$.
3. Question: Determine the work done by 1 mole of an ideal gas that expands isothermally at 300 K from 10 L to 20 L. Solution: Use $W=\mu RT\ln \frac{V_2}{V_1}$.

This summary and problem set is tailored for quick revision and application in NEET preparation. Review these key concepts regularly to reinforce understanding and accuracy in the exam.