Physics Q33 – JEE-MAIN 2026

Home Physics ThermodynamicsQuestion #506

Heat is supplied to a diatomic gas at constant pressure. Then the ratio of

Δ Q : Δ U : Δ W

is _______.

Clue 1 of 3: Degrees of Freedom

Recall the degrees of freedom for a diatomic gas.

🧠 Full Solution Path

Step 1: Identify Properties of Diatomic Gas and Constant Pressure Process✦ Active

For a diatomic gas, the degrees of freedom are $f = 5$ . The molar specific heat at constant volume is $C_{v} = \frac{f}{2} R = \frac{5}{2} R$ . The molar specific heat at constant pressure is $C_{p} = C_{v} + R = \frac{5}{2} R + R = \frac{7}{2} R$ .

Step 2: Express Thermodynamic Quantities in terms of

n R Δ T

○ Expand

The change in internal energy is given by $Δ U = n C_{v} Δ T$ . Substituting $C_{v}$ :

Δ U = n (\frac{5}{2} R) Δ T = \frac{5}{2} n R Δ T

For a constant pressure process, the work done is $Δ W = P Δ V$ . Using the ideal gas law, $P Δ V = n R Δ T$ :

Δ W = n R Δ T

The heat supplied at constant pressure is $Δ Q = n C_{p} Δ T$ . Substituting $C_{p}$ :

Δ Q = n (\frac{7}{2} R) Δ T = \frac{7}{2} n R Δ T

💡 Teacher's Secret Hint

Remember to use the correct specific heat capacities for the given process and gas type.

Step 3: Calculate the Ratio

Δ Q : Δ U : Δ W

○ Expand

Now, we find the ratio of these quantities:

Δ Q : Δ U : Δ W = \frac{7}{2} n R Δ T : \frac{5}{2} n R Δ T : n R Δ T

Dividing by $n R Δ T$ and multiplying by 2 to clear fractions:

\frac{7}{2} : \frac{5}{2} : 1 ⟹ 7 : 5 : 2

💡 Teacher's Secret Hint

Ensure all terms are expressed with a common factor before simplifying the ratio.

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Physics Question 33 – JEE-MAIN 2026

Clue 1 of 3: Degrees of Freedom

🧠 Full Solution Path