Question

Work done in an adiabatic process between a given pair of end states depends on
A. The end states only
B. Particular adiabatic process
C. Mass of the system
D. None of the above

Answer 1

Hint: End states of a system determined from the internal energy of the system. In an adiabatic system, there is no heat transferring. So, the dQ will be zero.
Formula used: \[dQ=dW+dU\], where dQ is the heat energy, dW is the work done and dU is the internal energy.

Complete Step-by-Step solution:
End states can be determined from the internal energy levels at those states. If we are considering a system S, the heat energy (dQ) can be used to do work (dW) or to store energy as potential energy (dU).
\[dQ=dW+dU\]
The interesting thing is that, for an adiabatic process dQ will be zero. So, the work done in an adiabatic process depends only on the internal energy. i.e. end states. Therefore the correct option is A.

Additional information:
If we are taking a real situation, it is difficult to build an adiabatic system. Since thermal isolation is a little bit hard to achieve. So the adiabatic process is an ideal case. However, we can approximate some situations as an adiabatic process. The situations are; fairly well thermally isolated systems, a fast process in which there is no time for the escape or incoming of heat and a very large system.
We can derive the equation of the adiabatic process.
According to the first law of thermodynamics, we can write as,
\[dQ=dW+dU\]
For an adiabatic process, dQ will be zero.
\[dW+dU=0\]
Where,
\[\begin{align}
  & dW=PdV \\
 & dU={{C}_{v}}dT \\
\end{align}\]
P is the pressure, dV is the change in volume, \[{{C}_{v}}\] is the specific heat at constant volume and dT is the change in temperature.
\[PdV+{{C}_{v}}dT=0\]…………………(1)
Now we can use the ideal gas equation.
\[PV=RT\]
After differentiating the ideal gas equation, we can write as,
\[PdV+VdP=RdT\]…………………….(2)
Substitute this equation into equation (1)
\[{{C}_{v}}\left[ \dfrac{PdV+VdP}{R} \right]+PdV=0\]
\[{{C}_{v}}\left[ PdV+VdP \right]+RPdV=0\]………….(3)
We can use Mayer’s relation here,
\[{{C}_{p}}-{{C}_{v}}=R\]
\[{{C}_{p}}=R+{{C}_{v}}\]
We can substitute this into the equation (3)
\[{{C}_{v}}VdP+{{C}_{p}}PdV=0\]
Divide this equation with \[P{{C}_{v}}V\]
\[\dfrac{dP}{P}+\dfrac{{{C}_{p}}}{{{C}_{v}}}\dfrac{dV}{V}=0\]
We can integrate this equation,
\[\gamma \ln V+\ln P=\ln C\], here \[\dfrac{{{C}_{p}}}{{{C}_{v}}}=\gamma \] known as adiabatic index and C is a constant.
We can write this equation as,
\[P{{V}^{\gamma }}=C\]

Note: In the adiabatic process, work done is due to the change in internal energy. Here the temperature can be varied but there is no transfer of heat. In the isothermal process, work done is due to the change in heat energy of the system. Here the temperature cannot be varied but there is the transfer of heat.