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(D) The container is rigid,which means the volume of the ideal gas remains constant,so the work done $W = P\Delta V = 0$.According to the First Law of

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$30$

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(D) The container is rigid,which means the volume of the ideal gas remains constant,so the work done $W = P\Delta V = 0$.According to the First Law of Thermodynamics,$\Delta Q = \Delta U + W$.Since $W = 0$,the change in internal energy is equal to the heat supplied: $\Delta U = \Delta Q$.The heat supplied by the filament is given by Joule's heating formula: $\Delta Q = I^2Rt$.Given: $I = 1 \,A$,$R = 100 \,\Omega$,and $t = 5 \,min = 5 	imes 60 \,s = 300 \,s$.Substituting the values: $\Delta U = (1)^2 	imes 100 	imes 300 = 30,000 \,J$.Converting to $kJ$: $\Delta U = 30 \,kJ$.

$A$ thermally insulated rigid container contains an ideal gas heated by a filament of resistance $100 \,\Omega$ through a current of $1 \,A$ for $5 \,min$. The change in internal energy is...... $kJ$.

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