Question

An ideal gas with adiabatic exponent $\gamma ,$is according to the law $P = \alpha v$where $\alpha$is a constant. The initial volume of the gas is ${V_0}.$ As a result increases $\eta$times. Find the increment in internal energy and work done.

Answer 1

An adiabatic process is the process where there is no loss or gain of heat during the entire process. Using the first law of thermodynamics the change in internal energy and the work done can be found out.

Step by step answer: No heat change occurs in adiabatic process but this process has a change in temperature.
If in a thermodynamic process work transfer of a system are functionless, there is no transfer of heat or of matter.
In the given problem we have $P = \alpha V$(1)
$P =$pressure of ideal gas
$V =$volume of ideal gas
$\alpha =$constant
We can write this equation as
$\Rightarrow \dfrac{P}{V} = \alpha$or
$\Rightarrow P{V^{ - 1}} = \alpha$ (2)
Compare this equation by $P = \alpha {V^m}$
We get, $m = - 1$
We have given
Initial volume${V_1} = {V_0}$
Final volume is $\eta$times of initial volume
$\therefore {V_2} = \eta {V_0}$
From equation (1)
Initial pressure ${P_1} = \alpha {V_0}$_______ (3)
Final pressure ${P_2} = \alpha \eta {V_0}$______ (4)
Change in internal energy is given by
$\Delta V = nRT$_______ (5)
Where $n =$number of moles
$R =$gas constant
$T =$absolute temperature
Since adiabatic exponent is $y$and temperature changes from ${T_1}$to ${T_2}.$
Therefore, the equation (5) becomes
$\Delta U = \eta \times \dfrac{R}{{(y - 1)}} \times ({T_2} - {T_1})$______ (6)
We know that $nRT = \Delta P\Delta V$for an ideal gas
Therefore, equation (6) can be written as
$\Rightarrow \Delta U = \dfrac{{{P_2}{V_2} - {P_1}{V_1}}}{{(y - 1)}}$_______ (7)
Substituting value of equation (3) and (4) we get
$\Rightarrow \Delta V = \dfrac{{[\alpha \eta {\gamma _0}(\eta {V_0}) - \alpha {V_0}({V_0})]}}{{y - 1}}$
$\Rightarrow \Delta V = \dfrac{{\alpha V_0^2({\eta ^2} - 1)}}{{y - 1}}$_______ (8)
Work done $w = \dfrac{{{P_2}{V_2} - {P_1}{V_1}}}{{1 - m}}$
$\therefore w = \dfrac{{\alpha V_0^2({\eta ^2} - 1)}}{{1 - m}}$ $m = - 1$
$\therefore w = \dfrac{{\alpha V_0^2({\eta ^2} - 1)}}{{1 - ( - 1)}}$
$\therefore w = \dfrac{{\alpha V_0^2({\eta ^2} - 1)}}{2}$

$\therefore$Work done by system$= \dfrac{{\alpha V_0^2({\eta ^2} - 1)}}{2}$

Note: Students may go wrong with the work done, while finding out work done using the first law of thermodynamics if work done obtained is positive it means work is done by the system, and if it is negative then work is done on the system.