Physics Q35 – JEE-MAIN 2026

Home Physics ThermodynamicsQuestion #515

If 2 mole of an ideal monoatomic gas at temperature

T

, is mixed with 6 mole of another ideal monoatomic gas at temperature

2 T

then the temperature of mixture is :

Clue 1 of 3: Conservation of Energy

When ideal gases are mixed without external heat exchange, the total internal energy of the system remains constant.

🧠 Full Solution Path

Step 1: Identify the Principle of Conservation of Internal Energy✦ Active

When two ideal gases are mixed without any heat exchange with the surroundings, the total internal energy of the system is conserved. This means the initial total internal energy equals the final total internal energy of the mixture.

Step 2: Apply the Formula for Internal Energy and Conservation○ Expand

For an ideal monoatomic gas, the internal energy $U = n C_{v} T$ , where $C_{v} = \frac{3}{2} R$ . Applying the conservation of internal energy:

U_{i n i t i a l} = U_{f i n a l}

n_{1} C_{v} T_{1} + n_{2} C_{v} T_{2} = (n_{1} + n_{2}) C_{v} T_{f}

Since both gases are monoatomic, $C_{v}$ is the same for both and for the mixture, so it cancels out:

n_{1} T_{1} + n_{2} T_{2} = (n_{1} + n_{2}) T_{f}

💡 Teacher's Secret Hint

Remember that $C_{v}$ is specific to the type of gas (monoatomic, diatomic, etc.). Since both are monoatomic, $C_{v}$ is constant and can be cancelled.

Step 3: Substitute Values and Solve for Final Temperature○ Expand

Given $n_{1} = 2$ mol, $T_{1} = T$ , $n_{2} = 6$ mol, $T_{2} = 2 T$ . Substitute these values into the equation:

(2) (T) + (6) (2 T) = (2 + 6) T_{f}

2 T + 12 T = 8 T_{f}

14 T = 8 T_{f}

T_{f} = \frac{14 T}{8}

T_{f} = \frac{7}{4} T

The final temperature of the mixture is $\frac{7}{4} T$ .

💡 Teacher's Secret Hint

Ensure careful algebraic manipulation to avoid calculation errors.

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Physics Question 35 – JEE-MAIN 2026

Clue 1 of 3: Conservation of Energy

🧠 Full Solution Path