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If in an isothermal process the volume of ideal gas is halved, then we can say that:
A. Internal Energy of the system decreases
B. Internal Energy of the system Increases
C. Work done by the gas is negative
D. Work done by the gas is positive


Answer
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Hint:
In the case, when a problem is based on thermodynamics, we know that both the parameters whether it is work done or internal energy vary with the given conditions of the system and surroundings. Hence, use the scientific formula of work done $dW = PdV$ at a constant temperature to state the answer for the given problem.

Formula used:
Boyles Law Equation, $PV = constant\,$
and, work done in an isothermal process, $Workdone = dW = PdV$


Complete step by step solution:
Since the internal energy is the function of temperature i.e., $U = f\left( T \right)$ . As the process is isothermal $T = constant$ , therefore, $U = constant$, and as a result of which Internal energy of the system neither increases nor decreases which means options (A) and (B) are incorrect.
Also, we know that at a constant temperature, the change in volume of a gas is inversely proportional to the pressure exerted by it (according to Boyles Law).
i.e., $PV = constant\, = k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\,(1)$
which is also called a constant temperature equation or Isothermal Process equation.
Therefore, in an isothermal, Work done is defined as: -
$Workdone = dW = PdV$

Thus, in an isothermal process if the volume of an ideal gas is halved, it means that the gas is getting compressed. Hence work is done on the system, which will be considered negative.
So, the work done by the gas is negative.


Hence, the correct option is (C) Work done by the gas is negative.





Note:
In this problem, to determine what will happen when the volume of gas is halved at a constant temperature, we will evaluate the work done by the gas by using the formula $dW = PdV$ and use $U = f\left( T \right)$ . Also, support your answer with proper reasons while writing an answer to this kind of conceptual problem.