Solution to 2001 Problem 36

(A) This is the definition of an adiabatic process.


\begin{align*}dS = \frac{dQ}{T}\end{align*}
Because the process is adiabatic, dQ = 0, so dS = 0, so the integral of dS over the entire process also equals 0.

(C) By the Second Law of Thermodynamics,

\begin{align*}\Delta U = Q - W\end{align*}
where \Delta U is the change in the internal energy of the gas, Q is the heat added to the gas, and W is the work done BY the gas. Since the process is adiabatic, Q = 0 and \Delta U = - W. W is given by

\begin{align*}W = \int P dV\end{align*}
\begin{align*}\Delta U = - W = - \int P dV\end{align*}

(D) As discussed in part (C),
\begin{align*}W = \int P dV\end{align*}

(E) The temperature of an ideal gas is related to the internal energy by
\begin{align*}U = 3/2 N k T\end{align*}
where N is the number of gas molecules, T is the temperature of the gas, and k is Boltzmann's constant. The internal energy of the gas is not constant (see part (C)), so the temperature cannot stay constant.

Therefore, answer (E) is correct.

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