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**Hint:**We are given an isothermal situation where an ideal gas is compressed. We know that when we compress a gas the volume of the gas gets decreased due to which the area also decreases. By finding the relation of this situation with pressure, we will get the solution (By using the ideal gas equation).

**Formula used:**

$PV=nRT$

**Complete answer:**

In the question it is said that when we compress an ideal gas isothermally, the pressure will increase. We are asked what causes the pressure to increase in such a situation.

We know that when we compress a gas isothermally, its temperature is constant and volume of the gas decreases.

Now let us consider the ideal gas equation. It is given as,

\[PV=nRT\], were ‘T’ is the temperature, ‘R’ is the ideal gas constant, ‘n’ is the number of moles, ‘V’ is the volume and ‘P’ is the pressure.

We know that in the given case the number of moles and temperature is a constant and ‘R’ is already a constant.

Therefore we can say that,

$PV=\text{constant}$

From the above equation we can say that pressure is inversely proportional to volume, i.e.

$P\propto \dfrac{1}{V}$

It simply says that as volume decreases, pressure increases.

Since in the given case it is said that we compress the volume; the volume of the gas decreases.

We know that as the volume decreases, area also decreases. So, when the area decreases the number of collisions per unit area increases.

This causes the increase in pressure.

**So, the correct answer is “Option C”.**

**Note:**

We can solve this using the kinetic theory of gases also.

We know that the average speed of a gas molecule at constant temperature is given by the equation,

${{V}_{avg}}=\sqrt{\dfrac{8KT}{\pi M}}$

From this equation we get,

$\Rightarrow {{V}_{avg}}=\sqrt{T}$

Thus the average velocity of each molecule increases and hence the number of collisions also increases, i.e. the number of collisions per unit area with the walls of the container increases.

Hence we get the solution.

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