N moles of a diatomic gas in a cylinder are a | vedclass

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(A) The initial internal energy of $N$ moles of diatomic gas is given by $U_i = N \left( \frac{5}{2} RT \right)$.When $n$ moles of diatomic gas dissoc

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$\frac{1}{2}nRT$

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(A) The initial internal energy of $N$ moles of diatomic gas is given by $U_i = N \left( \frac{5}{2} RT \right)$.When $n$ moles of diatomic gas dissociate into monoatomic gas,each diatomic molecule splits into two monoatomic atoms. Thus,$n$ moles of diatomic gas produce $2n$ moles of monoatomic gas.The remaining amount of diatomic gas is $(N-n)$ moles.The final internal energy $U_f$ is the sum of the energy of the monoatomic gas and the remaining diatomic gas:$U_f = (2n) \left( \frac{3}{2} RT \right) + (N-n) \left( \frac{5}{2} RT \right)$.$U_f = 3nRT + \frac{5}{2}NRT - \frac{5}{2}nRT = \frac{5}{2}NRT + \frac{1}{2}nRT$.The change in internal energy $\Delta U = U_f - U_i$ is:$\Delta U = \left( \frac{5}{2}NRT + \frac{1}{2}nRT \right) - \left( \frac{5}{2}NRT \right) = \frac{1}{2}nRT$.

Column-$I$	Column-$II$
$(a)$ Kinetic energy per unit mole of gas.	$(i)$ $\frac{1}{2}RT$
$(b)$ Kinetic energy per one molecule of gas.	$(ii)$ $\frac{3}{2}RT$
	$(iii)$ $\frac{3}{2}k_BT$

$N$ 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?

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Column-$I$ represents physical quantity and Column-$II$ represents formula. Match them correctly:
Column-$I$ Column-$II$
$(a)$ Kinetic energy per unit mole of gas. $(i)$ $\frac{1}{2}RT$
$(b)$ Kinetic energy per one molecule of gas. $(ii)$ $\frac{3}{2}RT$
$(iii)$ $\frac{3}{2}k_BT$

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