Question

Air expands from $5$ liters to $10$ liters at $2$ atm pressure, External work done is
(A) $10J$
(B) $1000J$
(C) $3000J$
(D) $300J$

Answer 1

Hint We know here that the air expands volume and we know the pressure so that we know the definition of the work performed and the type of the work performed, then we calculate the external work performed by using the specific formula to solve the problem.
Useful formula
Work done,
\[W\;{\text{ }} = \;{\text{ }}P{\text{ }}\left( {V2{\text{ }} - {\text{ }}V1} \right)\]
where,
\[\;V2{\text{ }} = \] final volume
\[V1{\text{ }} = \] initial volume.
as pressure $P$ is constant,

Complete step by step answer
Given by,
Initial velocity,\[V1 = 5liter = 5 \times {10^{ - 3}}{m^3}\]
Final velocity \[V2 = 10liter = 10 \times {10^{ - 3}}{m^3}\]
Pressure \[P = 2atm = 2 \times 105Pa\]
Work done:
The work performed is defined as the multiplication of displacement magnitude d and the component of the force in the direction of displacement.
Work performed on a system by external force is similar to the change in the system's total energy and the conversion of energy in the form of kinetic energy, but the system may transform it to some other form of energy instantaneously.
As we know that the formula for work done,
Here,
Work done, \[W = P\left( {V2 - V1} \right)\]
We understand that energy can neither be produced nor destroyed, so that energy has to be converted into some other form. In this scenario, this is referred to as job done. When negative energy is carried out the energy reduces and increases when good work is performed.
Substituting the given value in above equation
We get,
\[W = 2 \times 105\left( {10 - 5} \right) \times 10 - 3\]
On simplifying we get,
\[W = 2 \times 100 \times 5\]
Solving the above equation,
We get,
\[W = 1000J\]
Hence,

Thus, option B is the correct answer.

Note We should know that the air is thus composed of molecules that are constantly in motion. The molecules begin to vibrate and bump into each other while the air warms up, growing the space around each molecule. The air expands and becomes less compact, so each molecule requires more space for motion.