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If a gas X be confined inside a bulb as shown, by
what percent will the pressure of the gas be higher or lower than the atmospheric pressure? (Take the atmospheric pressure equal to 101.3 kPa)
a)4.75% higher
b)4.75% lower
c)6.75% higher
d)6.75% lower
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
If a gas X be confined inside a bulb as shown, by ... morewhat percent...
Explanation: Pa = Patm = 101.3
Pb = Pa + 0.9 * 9.81 * 0.03 = 101.56
Pc = Pb + 13.6 * 9.81 * 0.04 = 106.9
Pd = Pc – 1 * 9.81 * 0.05 = 106.41
Pe = Pd – 0.9 * 9.81 * 0.04 = 106.1
PX = Pe = 106.1
Since, PX > Patm, the percentage by which the pressure of the gas is higher than the atmospheric pressure will be
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Most Upvoted Answer
If a gas X be confined inside a bulb as shown, by ... morewhat percent...
Given: Atmospheric pressure = 101.3 kPa

To find: The percentage change in pressure of gas X confined inside a bulb

Solution:

We can use the Boyle's law to solve this problem. According to Boyle's law, the pressure of a gas is inversely proportional to its volume at a constant temperature.

P1V1 = P2V2

where P1 and V1 are the initial pressure and volume of the gas, and P2 and V2 are the final pressure and volume of the gas.

In this case, the gas X is confined inside a bulb, so its volume is constant. Let us assume that the initial pressure of the gas X is P1 and the final pressure is P2.

P1V = P2V

P1 = P2V/V

P1/P2 = V/V2

From the given diagram, we can see that the volume V2 is equal to the sum of the volumes of the bulb and the water in the flask.

V2 = V_bulb + V_water

We can assume that the volume of the water in the flask is negligible compared to the volume of the bulb, so we can write:

V2 ≈ V_bulb

Substituting this in the above equation, we get:

P1/P2 = V/V_bulb

P2/P1 = V_bulb/V

We know that the atmospheric pressure is 101.3 kPa, which is equivalent to the pressure of the air outside the bulb.

Let us assume that the initial pressure of the gas X is equal to the atmospheric pressure, i.e., P1 = 101.3 kPa.

The volume of the bulb is not given, so we cannot calculate the final pressure of the gas X directly. However, we can use the given information to estimate the percentage change in pressure.

From the diagram, we can see that the water level in the flask is at a height of 5 cm above the level of the water in the beaker. This means that the pressure of the water at the bottom of the flask is higher than the atmospheric pressure by an amount equal to the pressure due to the height of the water column.

We can calculate this pressure using the formula:

P = ρgh

where P is the pressure due to the height h of the water column, ρ is the density of water, and g is the acceleration due to gravity.

Substituting the given values, we get:

P = 1000 kg/m³ × 9.81 m/s² × 0.05 m ≈ 490 Pa

This pressure is equivalent to 0.49 kPa. Therefore, the final pressure of the gas X is:

P2 = 101.3 kPa + 0.49 kPa ≈ 101.79 kPa

The percentage change in pressure is:

ΔP/P1 × 100% = (P2 - P1)/P1 × 100%

ΔP/P1 × 100% = (101.79 - 101.3)/101.3 × 100%

ΔP/P1 × 100% ≈ 0.48%

Rounding off to two decimal places, we get:

ΔP/P1 × 100% ≈ 0.48% ≈ 0.5%

Therefore, the percentage change in pressure of the gas
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If a gas X be confined inside a bulb as shown, by ... morewhat percent will the pressure of the gas be higher or lower than the atmospheric pressure? (Take the atmospheric pressure equal to 101.3 kPa)a)4.75% higherb)4.75% lowerc)6.75% higherd)6.75% lowerCorrect answer is option 'A'. Can you explain this answer?
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