JEE MAIN - Physics (2017 - 9th April Morning Slot - No. 17)
N moles of a diatomic gas in a cylinder are at a temperature T. Heat is supplied to the cylinder such that the temperature remains constant but n moles of the diatomic gas get converted into monoatomic gas. What is the change in the total kinetic energy of the gas ?
$${1 \over 2}$$ nRT
0
$${3 \over 2}$$ nRT
$${5 \over 2}$$ nRT
Explanation
Initial kinetic energy of N mole of diatomic gas,
Ki = N$${5 \over 2}$$ RT
Kinetic energy of n mole of monoatomic gas
= n $${3 \over 2}$$ RT
When n mole of diatomic gas converted into monoatomic gas then remaining diatomic gas = (N $$-$$ n)
Find kinetic energy,
KF = (2m)$${3 \over 2}$$RT + (N $$-$$ n)$${5 \over 2}$$ RT
= $${1 \over 2}$$ nRT + $${5 \over 2}$$ NRT
$$\therefore\,\,\,$$ Change in kinetic energy,
$$\Delta $$K = Kf $$-$$ Ki = $${1 \over 2}$$ nRT
Ki = N$${5 \over 2}$$ RT
Kinetic energy of n mole of monoatomic gas
= n $${3 \over 2}$$ RT
When n mole of diatomic gas converted into monoatomic gas then remaining diatomic gas = (N $$-$$ n)
Find kinetic energy,
KF = (2m)$${3 \over 2}$$RT + (N $$-$$ n)$${5 \over 2}$$ RT
= $${1 \over 2}$$ nRT + $${5 \over 2}$$ NRT
$$\therefore\,\,\,$$ Change in kinetic energy,
$$\Delta $$K = Kf $$-$$ Ki = $${1 \over 2}$$ nRT
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