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Work Done in Reversible Adiabatic Process

One mole of a...

Question

One mole of a gas is subjected to two process AB and BC, one after the other as shown in the figure. BC is represented by $P V^{n}$ = constant. We can conclude that (where T = temperature, W = work done by gas, V = volume and U = internal energy).

A

T_{A} = T_{B} = T_{C}

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B

V_{A} < V_{B}, P_{B}, < P_{C}

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C

W_{A B} < W_{B C}

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D

U_{A} < U_{B}

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Solution

The correct option is A $T_{A} = T_{B} = T_{C}$
The process in which change in pressure and volume takes place at a constant temperature, is called a isothermal change. It may be noted that in such a change total amount of heat of the system does not remain constant
The heat moves out to keep the temperature constant. When the gas is expanded,heat is absorbed. This absorption is compensated by the transfer of heat from the surroundings, again keeping the temperature constant. Such a change is called an isothermal change.
The temperature is constant

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