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17. First law of thermodynamics is a law of conservation of energy. a. Write the expression for first law of thermodynamics. (1) b. Distinguish between isothermal change and adiabatic change on the basis of first law of thermodynamics. (2) c. Deduce the work done during isothermal expansion. (2) d. What is the expression for work done during isothermal expansion?

 1.a) The first law of thermodynamics for closed systems was originally induced from empirically observed evidence, however, it is now taken to be the definition of heat via the law of conservation of energy and the definition of work in terms of changes in the external parameters of a system.

For a given system we let ?E kin = large-scale mechanical energy, ?E pot = large-scale potential energy, and ?E tot = total energy. The first two quantities are specifiable in terms of appropriate mechanical variables, and by definition
E^{\mathrm{tot}}=E^{\mathrm{kin}}+E^{\mathrm{pot}}+U\,\,.
For any finite process, whether reversible or irreversible,
\Delta E^{\mathrm{tot}}=\Delta E^{\mathrm{kin}}+\Delta E^{\mathrm{pot}}+\Delta U\,\,.
The first law in a form that involves the principle of conservation of energy more generally is
\Delta E^{\mathrm{tot}}=Q+W\,\,.
Here Q and W are heat and work added, with no restrictions as to whether the process is reversible, quasistatic, or irreversible.
 1.b) Adiabatic process: no heat is transferred.
Isothermal process: occurs at a constant temperature.

It might appear that temperature cannot change if no heat is added, but that is not the case. You can increase the temperature of a gas by compressing it; even if the piston and cylinder are insulated against heat transfer. Essentially, the compression process takes mechanical work and converts it to both an increase in pressure and an increase in temperature.

You can increase the temperature of a gas in an adiabatic process if you do some mechanical work on the gas.
2.c)

In thermodynamics, the work involved when a gas changes from state A to state B is simply

W_{A\to B} = \int_{V_A}^{V_B}p\,dV

For an isothermal, reversible process, this integral equals the area under the relevant pressure-volume isotherm, and is indicated in blue in the figure (at the bottom right-hand of the page) for an ideal gas. Again, p = nRT / V applies and with T being constant (as this is an isothermal process), we have:

W_{A\to B} = \int_{V_A}^{V_B}p\,dV = \int_{V_A}^{V_B}\frac{nRT}{V}dV = nRT\int_{V_A}^{V_B}\frac{1}{V}dV = nRT\ln{\frac{V_B}{V_A}}
for many systems, if the temperature is held constant, the internal energy of the system also is constant, and so \Delta U = 0. From First law of thermodynamics \Delta U=Q+W, so it follows that -Q =W for this same isothermal process.

When no heat flows into or out of the gas because the temperature is constant, then there is no work done. Thus, work=0 which means external pressure is zero. This is called free expansion


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