Exams › NEET › Physics

Change in internal energy from A → B: ΔU = f/2 nRΔT = f/2 nR (Tf - Ti)

1. f/2 {PVf - PVi}
2. f/2 {2 × 10³ × 6 - 5 × 10³ × 4}
3. f/2 {12 - 20} × 10³ J = 5 × (-4) × 10³ J
4. ΔU = -20 KJ

Correct answer: f/2 {12 - 20} × 10³ J = 5 × (-4) × 10³ J

Solution

The internal energy change is calculated using ΔU = f/2 nRΔT, which simplifies to f/2 {Tf - Ti}. Substituting the given values, the correct calculation is shown in option C.

Related NEET Physics questions

• If for a gas \( \frac{\boldsymbol{R}}{\boldsymbol{C}_{\boldsymbol{v}}}=\mathbf{0 . 6 7}, \) then the gas is made up of molecules which are :
• Use of thermometer is based on which law of thermodynamics?
• The total kinetic energy of 1 mole of \( N_{2} \) at \( 27 \mathrm{C} \) will be approximately
• The quantity \( \frac{2 U}{f k T} \) represents (where \( U= \) internal energy of gas
• A cyclic heat engine does \( 50 \mathrm{kJ} \) of work per cycle. If efficiency of engine is \( 75 \% \) the heat rejected per cycle will be:
• Thermistor material is pressed

⚔️ Practice NEET Physics free + battle 1v1 →