Exams › NEET › Physics

According to question work done by gas is given -150J so that according to it answer will be (d).

1. T1 = 273 + 27 = 300K
2. T2 = 273 + 927 = 1200K
3. For adiabatic process, P1(T1^(γ-1)) = P2(T2^(γ-1))
4. R1/R2 = (1200/300)^1.4 = 4^1.4

Correct answer: R1/R2 = (1200/300)^1.4 = 4^1.4

Solution

The work done by the gas is negative, indicating compression. The ratio of temperatures (T2/T1) raised to the power of γ (1.4 for diatomic gases) gives the correct ratio of R1/R2 as 4^1.4.

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 →