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In a thermodynamic process pressure of a fixed mass of a gas is changed in such a manner that the gas releases \(30 \mathrm{~J}\) of heat and \(10 \mathrm{~J}\) of work was done on the gas. If the initial internal energy of the gas was \(10 \mathrm{~J}\), then the final internal energy will be
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The correct answer is:
\(10 \mathrm{~J}\)
Given, heat, \(Q=-30 \mathrm{~J}\)
Work done on the gas, \(W=-10 \mathrm{~J}\)
Initial internal energy, \(U_i=-10 \mathrm{~J}\)
Final internal energy, \(U_f=\) ?
According to first law of thermodynamics,
\(\begin{array}{ll}
& Q=W+\Delta U \\
\Rightarrow & -30=-10+U_i-U_f \\
\Rightarrow & -30=-10-10-U_f \\
\Rightarrow & 30=10+10+U_f \\
\Rightarrow & 30-20=U_f \\
\Rightarrow & U_f=10 \mathrm{~J}
\end{array}\)
Work done on the gas, \(W=-10 \mathrm{~J}\)
Initial internal energy, \(U_i=-10 \mathrm{~J}\)
Final internal energy, \(U_f=\) ?
According to first law of thermodynamics,
\(\begin{array}{ll}
& Q=W+\Delta U \\
\Rightarrow & -30=-10+U_i-U_f \\
\Rightarrow & -30=-10-10-U_f \\
\Rightarrow & 30=10+10+U_f \\
\Rightarrow & 30-20=U_f \\
\Rightarrow & U_f=10 \mathrm{~J}
\end{array}\)
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