The vapour pressure of water at 20°C is 17.54 mmHg. What will be the vapour pressure of thewater in the apparatus shown after the piston is lowered, decreasing the volume of the gasabove the liquid to one half of its initial volume (assume temperature constant).(A) 8.77 mmHg (B) 17.54 mmHg(C) 35.08 mmHg (D) between 8.77 and 17.54 mmHg

To determine the vapor pressure of water in the apparatus after the piston is lowered, we can use the relationship between vapor pressure and volume.

According to the problem, the initial volume of the gas above the liquid is decreased to one half of its initial volume. Since the temperature is constant, we can assume that the amount of water vapor remains the same.

The relationship between vapor pressure and volume is known as Boyle's Law, which states that at constant temperature, the product of the initial pressure and volume is equal to the final pressure and volume. Mathematically, this can be expressed as:

P1 * V1 = P2 * V2

Where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.

In this case, we know that the initial pressure (P1) is 17.54 mmHg and the final volume (V2) is half of the initial volume. Let's denote the final pressure as P2.

Using the equation, we can rewrite it as:

17.54 mmHg * V1 = P2 * (0.5 * V1)

Simplifying the equation, we get:

17.54 mmHg = P2 * 0.5

To find the value of P2, we can rearrange the equation:

P2 = 17.54 mmHg / 0.5

P2 = 35.08 mmHg

Therefore, the vapor pressure of the water in the apparatus after the piston is lowered will be 35.08 mmHg.

So, the correct answer is (C) 35.08 mmHg.