Initial pressure and volume of a monoatomic ideal gas are P and V . The change in internal energy of this gas in adiabatic expansion to volume V_ final =27V is ________ J.

Question

Quantrex Academy · Accepted Answer

Correct answer: C. -(4)/(3)PV

For a monoatomic ideal gas, the ratio of specific heats is

\gamma = \dfrac{5}{3}

.

In an adiabatic process,

PV^{\gamma} = \text{constant}

.

P_1 V_1^{\gamma} = P_2 V_2^{\gamma}

P V^{5/3} = P_2 (27V)^{5/3}

P_2 = P \left(\dfrac{1}{27}\right)^{5/3} = \dfrac{P}{243}

The change in internal energy is given by:

\Delta U = \dfrac{P_2 V_2 - P_1 V_1}{\gamma - 1}

Substituting the values:

\Delta U = \dfrac{\left(\dfrac{P}{243}\right)(27V) - PV}{\dfrac{5}{3} - 1}

\Delta U = \dfrac{\dfrac{PV}{9} - PV}{\dfrac{2}{3}}

\Delta U = \dfrac{-\dfrac{8}{9}PV}{\dfrac{2}{3}} = -\dfrac{4}{3}PV

Answer:

-\dfrac{4}{3}PV

Initial pressure and volume of a monoatomic ideal gas are P and V ....

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