A gas is kept in a container having walls which are thermally non-conducting. Initially the gas has a volume of 800 cm^3 and temperature 27°C. The change in temperature when the gas is adiabatically compressed to 200 cm^3 is: (Take γ = 1.5; γ is the ratio of specific heats at constant pressure and at constant volume)

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q34 – JEE-MAIN 2025

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Home Physics ThermodynamicsQuestion #527

A gas is kept in a container having walls which are thermally non-conducting. Initially the gas has a volume of

800 {cm}^{3}

and temperature

27^{\circ} C

. The change in temperature when the gas is adiabatically compressed to

200 {cm}^{3}

is: (Take

γ = 1.5

;

γ

is the ratio of specific heats at constant pressure and at constant volume)

Clue 1 of 3: Adiabatic Process

An adiabatic process is one where no heat is exchanged with the surroundings.

🥷

Ninja Strategy•Careful Reading of the Question

Distinguish between the final temperature and the change in temperature to avoid selecting an option that only represents the final state.

🧠 Full Solution Path

Step 1: Identify Initial Conditions and Process Type✦ Active

The initial temperature is $T_{1} = 27^{\circ} C$ . Convert this to Kelvin: $T_{1} = 27 + 273 = 300 K$ . The initial volume is $V_{1} = 800 {cm}^{3}$ . The final volume is $V_{2} = 200 {cm}^{3}$ . The ratio of specific heats is $γ = 1.5$ . The process is adiabatic.

Step 2: Apply Adiabatic Relation to Find Final Temperature○ Expand

For an adiabatic process, the relationship between temperature and volume is $T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1}$ . We need to find the final temperature $T_{2}$ .

T_{2} = T_{1} {(\frac{V_{1}}{V_{2}})}^{γ - 1}

Substitute the given values: $T_{2} = 300 K \times {(\frac{800 {cm}^{3}}{200 {cm}^{3}})}^{1.5 - 1} = 300 K \times (4)^{0.5} = 300 K \times 2 = 600 K$ .

💡 Teacher's Secret Hint

Remember to convert temperature to Kelvin before using thermodynamic equations.

Step 3: Calculate the Change in Temperature○ Expand

The question asks for the change in temperature, $Δ T = T_{2} - T_{1}$ .

Δ T = 600 K - 300 K = 300 K

💡 Teacher's Secret Hint

Pay close attention to whether the question asks for the final temperature or the change in temperature.

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Physics Question 34 – JEE-MAIN 2025

Clue 1 of 3: Adiabatic Process

🧠 Full Solution Path