Isothermal Process

Solve by difficulty

If $\Delta U$ and $\Delta W$ represent the increase in internal energy and work done by the system respectively in a thermodynamical process, which of the following is true?

$\Delta U=-\Delta W,$ in an adiabatic process

$\Delta U=\Delta W,$ in an isothermal process

$\Delta U=\Delta W,$ in an adiabatic process

$\Delta U=-\Delta W,$ in an isothermal process

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In thermodynamic processes, which of the following statements is not true?

In an isochoric process, pressure remains constant

In an isothermal process the temperature remains constant

In an adiabatic process, process $PV^{\gamma}=constant$

In an adiabatic process, the system is insulated from the surroundings

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Concepts Covered - 1

Isothermal Process

Isothermal process - When a thermodynamic system undergoes a thermodynamic process in such a way that its temperature remains constant, then that process is called an Isothermal process.

So, T = constant and $\Delta T = 0$ .

Trick to recognize isothermal process -

The walls of the container must be perfectly conducting (no resistance) which allows the exchange of heat between the gas and surroundings.
The process of compression or expansion should be infinitely slow so that the process gets proper time for the exchange of heat.

Equation of isothermal process -

As we know that the equation of state is given by - $PV= nRT$

If T = constant and for a particular amount of gas ‘n’ is also constant.

So we can write that - $P.V = Constant$

Points in graph of isothermal process -

i) Curves obtained on P-V graph are called isotherms and the graphs are hyperbolic in nature.

ii) Slope of isothermal curve :

By differentiating PV = C

$P d V+V d P=0 \Rightarrow P d V=-V d P \Rightarrow \frac{d P}{d V}=-\frac{P}{V}$

$tan\theta = \frac{dP}{dV} = \frac{-P}{V}$

iii) The area between the isothermal curve and volume axis represents the work done in the isothermal process.

The formula of Work done in the isothermal process -

$\begin{array}{l}{W=n R T \log _{e}\left(\frac{V_{f}}{V_{i}}\right)=2.303n R T \log _{10}\left(\frac{V_{f}}{V_{i}}\right)} \\ {W=n R T \log _{e}\left(\frac{P_{i}}{P_{f}}\right)=2.303 nR T \log _{10}\left(\frac{P_{i}}{P_{f}}\right)}\end{array}$