An insulated container contains 4 , text{mole | vedclass

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(B) Initial state: $4 \, 	ext{moles}$ of diatomic gas at temperature $T$. The internal energy is $U_i = n C_v T = 4 	imes \frac{5}{2} RT = 10 RT$.Fi

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$Q = RT$

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(B) Initial state: $4 \, 	ext{moles}$ of diatomic gas at temperature $T$. The internal energy is $U_i = n C_v T = 4 	imes \frac{5}{2} RT = 10 RT$.Final state: $2 \, 	ext{moles}$ of diatomic gas remain, and $2 \, 	ext{moles}$ of diatomic gas dissociate into $4 \, 	ext{moles}$ of monoatomic gas. The temperature remains $T$.The internal energy of the final state is $U_f = (n_{di} C_{v,di} T) + (n_{mono} C_{v,mono} T) = (2 	imes \frac{5}{2} RT) + (4 	imes \frac{3}{2} RT) = 5 RT + 6 RT = 11 RT$.By the first law of thermodynamics, $Q = \Delta U = U_f - U_i = 11 RT - 10 RT = RT$.

An insulated container contains $4 \, \text{moles}$ of an ideal diatomic gas at temperature $T$. Heat $Q$ is supplied to this gas, due to which $2 \, \text{moles}$ of gas are dissociated into atoms, but the temperature of the gas remains constant. Then:

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