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NVAEasyJEE 2025Degrees of Freedom & Law of Equipartition

JEE Physics 2025 Question with Solution

The internal energy of air in 4 m×4 m×3 m4 \, m \times 4 \, m \times 3 \, m sized room at 11 atmospheric pressure will be _____ \times 10^6 \, J. (Consider air as a diatomic molecule)

Answer

Correct answer:12.2

Step-by-step solution

Standard Method

Given: Room size is 4 m×4 m×3 m4 \, \text{m} \times 4 \, \text{m} \times 3 \, \text{m} and air is treated as a diatomic ideal gas at 1 atm1\,\text{atm}.

Find: The internal energy of air in the room.

For a diatomic gas,

U=nCvTU = nC_vT

with

Cv=52RC_v = \frac{5}{2}R

Using PV=nRTPV = nRT,

n=PVRTn = \frac{PV}{RT}

So,

U=nCvT=PVRT⋅52RT=52PVU = nC_vT = \frac{PV}{RT} \cdot \frac{5}{2}RT = \frac{5}{2}PV

First calculate the room volume:

V=4×4×3=48 m3V = 4 \times 4 \times 3 = 48 \, \text{m}^3

Now substitute P=1.013×105 PaP = 1.013 \times 10^5 \, \text{Pa} and V=48 m3V = 48 \, \text{m}^3:

U=52×(1.013×105)×48U = \frac{5}{2} \times (1.013 \times 10^5) \times 48 U=1.2156×107 JU = 1.2156 \times 10^7 \, \text{J} U=12.16×106 J≈12.2×106 JU = 12.16 \times 10^6 \, \text{J} \approx 12.2 \times 10^6 \, \text{J}

Therefore, the required numerical value is 12.212.2. The solution also shows 1212 in one place due to using the approximation 1 atm≈105 Pa1\,\text{atm} \approx 10^5\,\text{Pa}, but the detailed working concludes 12.212.2.

Using the direct thermodynamic relation

Given: Air is diatomic and occupies a room of volume 48 m348 \, \text{m}^3 at about 11 atmospheric pressure.

Find: Total internal energy.

The hint and first approach use the direct relation for a diatomic ideal gas:

U=52PVU = \frac{5}{2}PV

If 1 atm≈105 Pa1\,\text{atm} \approx 10^5\,\text{Pa} is used, then

U=52×105×48=12×106 JU = \frac{5}{2} \times 10^5 \times 48 = 12 \times 10^6 \, \text{J}

If the more accurate value 1.013×105 Pa1.013 \times 10^5\,\text{Pa} is used, then

U≈12.16×106 J≈12.2×106 JU \approx 12.16 \times 10^6 \, \text{J} \approx 12.2 \times 10^6 \, \text{J}

Therefore, the numerical value from the detailed solution is 12.212.2.

Common mistakes

  • Using U=32PVU = \frac{3}{2}PV for air. That expression is for a monoatomic ideal gas, so it underestimates the internal energy. Since air is treated as diatomic here, use U=52PVU = \frac{5}{2}PV instead.

  • Calculating the room volume incorrectly. Multiplying the dimensions wrongly changes the final answer directly. First find V=4×4×3=48 m3V = 4 \times 4 \times 3 = 48 \, \text{m}^3 and then substitute.

  • Ignoring the pressure unit conversion. Atmospheric pressure must be written in pascal before using PVPV in SI units. Use either 1 atm≈105 Pa1\,\text{atm} \approx 10^5\,\text{Pa} or 1.013×105 Pa1.013 \times 10^5\,\text{Pa} consistently.

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