JEE MAIN - Physics (2020 - 4th September Morning Slot - No. 22)
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 _______.
Answer
266
Explanation
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, $$\Delta $$U = nCv$$\Delta $$T
$$ \therefore $$ $$\left( {0.1 \times {{3R} \over 2} \times 200} \right) + \left( {0.05 \times {{3R} \over 2} \times 400} \right)$$ = $$\left( {0.15 \times {{3R} \over 2} \times {T_f}} \right)$$
$$ \Rightarrow $$ (20 + 20) = 0.15 Tf
$$ \Rightarrow $$ Tf = 266.67
$$ \therefore $$ Total internal energy of gases remain same.
u1 + u2 = u1' + u2'
We know, $$\Delta $$U = nCv$$\Delta $$T
$$ \therefore $$ $$\left( {0.1 \times {{3R} \over 2} \times 200} \right) + \left( {0.05 \times {{3R} \over 2} \times 400} \right)$$ = $$\left( {0.15 \times {{3R} \over 2} \times {T_f}} \right)$$
$$ \Rightarrow $$ (20 + 20) = 0.15 Tf
$$ \Rightarrow $$ Tf = 266.67
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