Question

The relation between internal energy $U$ , pressure $P$ and volume $V$ of a gas in an adiabatic process is $U = a + bPV$ , where $a$ and $b$ are positive constants. What is the value of $\gamma$ ?
(A) $\dfrac{a}{b}$
(B) $\dfrac{{b + 1}}{b}$
(C) $\dfrac{{a + 1}}{a}$
(D) $\dfrac{b}{a}$

Answer 1

Hint : Use the definition of adiabatic process , the first law of thermodynamics and the relation of pressure and volume then use to find the value of the constant. In adiabatic processes total heat exchanged is always zero. The first law of thermodynamics states that the total exchange in heat is used by a system in two ways. First one is to increase its internal energy and rest is used to do some external work. $dQ = dU + dW$ . Adiabatic relation of a gas is given by, $P{V^\gamma } = K$ where, $K$ is a constant and $\gamma$ is the ratio of molar specific heats of the gas.

Complete Step By Step Answer:
We know for an adiabatic process net exchange of heat is always zero. Hence, for an adiabatic process $dQ = 0$ .
Now we know, the first law of thermodynamics states that the total exchange of heat is used by a system in two ways. First one is to increase its internal energy and rest is used to do some external work. $dQ = dU + dW$ .
Hence, from the first law of thermodynamics we can write, $0 = dU + PdV$ where, we know work done $dW = PdV$ .
Now, we have given, $U = a + bPV$ . Hence, differentiating we get,
$dU = d\left( {a + bPV} \right)$
Or, $bPdV + bVdP$ .
Now, putting this in the First law we get,
$bPdV + bVdP + pdV = 0$
Or, $\left( {b + 1} \right)PdV + bVdP = 0$
On further simplifying we get,
$\left( {b + 1} \right)\dfrac{{dV}}{V} + \dfrac{{dP}}{P} = 0$
On integrating we get,
$\left( {b + 1} \right)\log V + b\log P = k$ Where, $k$ is a constant
Or, ${V^{b + 1}}{p^b} = k$
Or, $P{V^{\dfrac{{b + 1}}{b}}} = k$
Now, we know for an adiabatic process the relation between $P$ and $V$ is given by, $P{V^\gamma } = K$ where, $K$ is a constant and $\gamma$ is the ratio of molar specific heats of the gas.
Hence, comparing these two equations we get,
$\gamma = \dfrac{{b + 1}}{b}$ .
Hence, the value of $\gamma$ is $\dfrac{{b + 1}}{b}$ .
Hence, option ( B ) is correct.

Note :
The work done by a gas is always equal to the area under the $PV$ -curve. So, we write for an infinitesimal work done $dW = PdV$ . Where, $P$ is a function of $V$ and $T$ . Total work done by the gas is the integration of it $W = \int {P(V,T)dV}$ .