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(D) Since the container is rigid,the volume of the ideal gas remains constant,so the work done $W = P \Delta V = 0$.According to the first law of ther

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$30$

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(D) Since the container is rigid,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 $W = 0$,we have $\Delta Q = \Delta U$.The heat supplied by the filament is given by Joule's law of heating: $\Delta Q = i^2 R t$.Given: $i = 1 \, A$,$R = 100 \, \Omega$,and $t = 5 \, 	ext{min} = 5 	imes 60 \, 	ext{s} = 300 \, 	ext{s}$.Substituting the values: $\Delta U = (1)^2 	imes 100 	imes 300 = 30,000 \, 	ext{J} = 30 \, 	ext{kJ}$.

An ideal gas is contained in a thermally insulated rigid container. It is heated by a filament of $100 \, \Omega$ resistance by passing a current of $1 \, A$ for $5 \, \text{min}$. The change in internal energy is .... $kJ$?

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