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QJEEM20Ph63ca: JEE Main 2020 Physics - Class 11 JEEM20Ph63ca Solution: A closed vessel contains 0.1 mole of a monoatomic ideal gas at…

A closed vessel contains 0.1 mole of a monoatomic ideal gas at 200 K. If 0.05 mole of the same gas at 400 K is added to it, the final equilibrium temperature (in K) of the gas in the vessel will be close to _______.

Step-by-Step Verified Solution

As work done on gas and heat supplied to the gas are zero,

∴\therefore Total internal energy of gases remain same.

u1 + u2 = u1' + u2'

We know, Δ\Delta U = nCv Δ\Delta T

∴\therefore (0.1×3R2×200)+(0.05×3R2×400)\left( {0.1 \times {{3R} \over 2} \times 200} \right) + \left( {0.05 \times {{3R} \over 2} \times 400} \right) = (0.15×3R2×Tf)\left( {0.15 \times {{3R} \over 2} \times {T_f}} \right)

⇒\Rightarrow (20 + 20) = 0.15 Tf

⇒\Rightarrow Tf = 266.67

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