Question

A gas does $8J$ of external work during adiabatic expression. Its temperature falls by $4K$. The internal energy of the gas will be:
A. increased by $8J$
B. decrease by $4J$
C. decrease by $8J$
D. increase by $2J$

Answer 1

Hint: A gas can do work through various thermodynamic processes. For finding the change in internal energy of the gas system, we will use the equation of adiabatic thermodynamic processes and the first law of thermodynamics to relate the terms; energy transfer, work done, and internal energy of the gas system.

Formula used:
First law of thermodynamics:
$dQ=dU+dW$

Complete step by step answer:
Thermodynamic process is the passage of a thermodynamic system from an initial state to a final state of thermodynamic equilibrium. The initial and the final states of a system are the defining elements of a thermodynamic process. The four types of thermodynamic process are: Isothermal, Isochoric, Isobaric and Adiabatic.
In an isothermal process, the temperature of a system remains constant. Thermal equilibrium is maintained during the process. Equation of isothermal process is given by,
 $PV=\text{ constant}$
In an isochoric process, the volume of the closed system undergoing the process remains unchanged. The Isochoric process is an Isovolumetric process. Equation of Isochoric process is given by,
$\dfrac{P}{T}=\text{ constant}$
In the Isobaric process, the pressure of the system stays constant. Equation of Isobaric process is given by,
$\dfrac{V}{T}=\text{ constant}$
In adiabatic processes, the heat and mass transfer between system and surroundings remains zero. Equation of adiabatic process is given by,
$P{{V}^{\gamma }}=\text{ constant}$
Where,
$P$ is the pressure of the system
$V$ is the volume of the system
$T$ is the temperature of the system
$\gamma $ is the ratio of specific heats $\left( \gamma =\dfrac{{{C}_{P}}}{{{C}_{V}}} \right)$
The first law of thermodynamics is a version of the law of conservation of energy, adapted for thermodynamic processes. This law distinguishes two types of transfer of energy: energy in the form of heat and energy in the form of work done. The law relates the two types of energy to the body's state function, internal energy.
Expression for first law of thermodynamics:
$dQ=dU+dW$
Where,
$dQ$ is the heat transfer between system and surroundings
$dU$ is the change in internal energy of the system
$dW$ is the work done on the system or by the system
From the first law of thermodynamics,
$dQ-dU=dW$
For an adiabatic process,
$dQ=0$
Therefore,
$dU=-dW$
$dU=-8J$
The internal energy of the gas decreases by $8J$
Hence, the correct option is C.

Note:
Students need to remember the equations of thermodynamic process and the work done expression for them to solve the above equation. Also, keep in mind that in an Adiabatic process, heat and mass transfer between system and surroundings remain zero. Therefore, according to the first law of thermodynamics, when a gas undergoes adiabatic process, internal energy of the system is equal to the work done by the system with a negative sign.