In an isothermal process if heat is released from an ideal gas then
A. the internal energy of the gas will increase.
B. the gas will do positive work.
C. the gas will do negative work.
D. the given process is not possible.

Hint: An isothermal process, is a thermodynamic process in which temperature remains constant. Applying the condition for ideal gas in the formula of internal energy, we will get the change in internal energy in an isothermal process and then we can substitute it in the differential form of thermodynamics.

Formula used:
\[\begin{align}
  & dU=n{{C}_{v}}dT \\
 & dQ=dU+dW \\
\end{align}\]

Complete step by step answer:
The change in internal energy of a system can be given as
\[dU=n{{C}_{v}}dT\]
Where n is amount of substance, \[{{C}_{v}}\] is the molar specific heat and dT is the change in temperature.
In case of isothermal process, there is no change in temperature for an ideal gas
\[\begin{align}
  & \Rightarrow dT=0 \\
 & \Rightarrow dU=0 \\
\end{align}\]
That means for an isothermal process there will be no change in the internal energy.
Now considering the differential form of the first law of thermodynamic
\[dQ=dU+dW\]
Here as \[dU=0\], therefore
\[dQ=dW\]
Which implies that the heat absorb or release is equal to the work done.
As per the question heat is released there dQ will be negative and hence dW will also be negative.

So, the correct answer is “Option C”.

Note:
In case heat was absorbed then the gas would have done positive work. In an isothermal process the internal energy will remain constant always. Although in actual practice we can’t convert the whole energy in work, always some heat is either dissipated or absorbed or created by the system.