Physics College

The saturated vapour pressure at 20°C and 8°C is 17.54mmHg and 8.05mmHg respectively. Find,

(i) Relative humidity at the room temperature.
(ii) How much mass of water should evapourate such that it saturate at 20°C (1mmHg = 133.3Pa)​

The Saturated Vapour Pressure At 20C And 8C Is 17.54mmHg And 8.05mmHg Respectively. Find,(i) Relative

Answers

Answer 1

1.) The relative humidity is 100 percent. 2.)To saturate the air at 20°C, 0.018g of water should evaporate.

(i) To calculate the relative humidity at room temperature, compare the amount of water vapour in the air to the maximum amount of water vapour that the air can hold at that temperature.

The saturated vapour pressure at that temperature determines the maximum amount. The relative humidity is stated as a percentage and can be calculated using the formula:

(Actual vapour pressure / Saturated vapour pressure) times 100% = Relative humidity

The saturation vapour pressure at room temperature (20°C) is 17.54mmHg. The relative humidity is 100% if the actual vapour pressure in the air equals the saturation vapour pressure.This signifies that the air has held the most water vapour it can at that temperature.

(ii) Using the ideal gas law, we can calculate the amount of water that should evaporate to saturate the air at 20°C. PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin, according to the ideal gas equation.

This equation can be rearranged to calculate the amount of moles of water:

n = (PV) / (RT)

We know that the required pressure is 17.54mmHg, which is equal to 2.338 kPa. We also know that the temperature is 20° C, which is 293.15 K.The volume is not specified, but we can assume it is constant and hence ignore it. The gas constant is 8.31 joules per mol-K.

n = ((2.338 kPa) / (8.31 J/mol-K * 293.15K))

To calculate the mass of water, multiply the number of moles by the molar mass of water, which is about 18.015 g/mol.

0.01796 g = 0.000997 mol * 18.015 g/mol mass of water

To saturate the air at 20°C, around 0.018g of water should evaporate.

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Related Questions

A 40kg child is on a swing. At the bottom of a swing, the child attains a speed
of 4.0 meters per second. If the chain holding the swing is 4.0m long, what is
the Force of Contact by the chain on the child at the bottom of the swing
(assume the chain has no mass)?

Answers

The force of contact by the chain on the child at the bottom of the swing is 160 N. This is the minimum force required to keep the child moving in a circular path with a constant speed of 4.0 m/s. Any lesser force and the child will move away from the circular path and slow down.

At the bottom of the swing, the child has a velocity of 4.0 m/s and is being pulled downwards by the force of gravity. The tension in the chain holding the swing provides the necessary centripetal force to keep the child moving in a circular path. The centripetal force required is given by the formula F = (mv^2)/r, where F is the force, m is the mass of the child, v is the velocity, and r is the radius of the circular path. In this case, the radius is equal to the length of the chain, which is 4.0 m.

Substituting the given values, we get:

F = (40 kg x (4.0 m/s)^2)/4.0 m

F = 160 N

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The surface of the exit road is horizontal, not banked. (See figure.) If the static
friction between the tires and the surface of the road is us = 0.688 and the
maximum speed with which the car can exit the highway safely without sliding is
25.2 m/s, what is the radius of curvature of a highway exit, r?

Answers

The radius of curvature of the highway exit is approximately 220 km.

To find the radius of curvature of the highway exit, we can use the centripetal force equation:

F = mv^2 / r

where F is the maximum static friction force, m is the mass of the car, v is the maximum safe speed, and r is the radius of curvature.

We can solve for r by rearranging the equation:

r = mv^2 / F

Substituting the given values, we have:

r = (1000 kg)(25.2 m/s)^2 / (0.688)

r = 2.20 x 10^5 m

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When 119 J of work is done in pushing a box horizontally 17 m, how much force is applied?

Answers

If 119 J of work is done in pushing a box horizontally 17 m, we can use the work-energy principle to find the force applied. The work-energy principle states that the work done on an object is equal to the change in its kinetic energy. Since the box is initially at rest, the work done on it is equal to its final kinetic energy, which is given by:

(1/2)mv^2

where m is the mass of the box and v is its final velocity. Since the box is pushed horizontally, the force applied is in the same direction as the displacement, so the work done is given by:

W = Fd

where F is the force applied and d is the displacement. Setting these two expressions equal to each other, we get:

Fd = (1/2)mv^2

Solving for F, we get:

F = (1/2)(m/d)v^2

We are not given the mass of the box or its final velocity, but we can use the fact that the work done is 119 J and the displacement is 17 m to find the force applied. Plugging in the given values, we get:

F = (1/2)(119 J / 17 m) = 3.5 N

Therefore, the force applied is 3.5 N.

. The force of gravity on the Moon is said to be one-sixth of that on the Earth. What would a mass of 12 kg weigh; (a) on the Earth


​

Answers

A. The weight of the mass on the earth is 117.6 N

A. The weight of the mass on the moon is 19.56 N

How do i determine the weight of the mass?

Weight is defined as follow:

Weight (W) = mass (m) × Acceleration due to gravity (g)

W = mg

Now we shall determine the weight. Details below:

A. Weight on earth

Mass (m) = 12 KgAcceleration due to gravity on earth (g) = 9.8 m/s²   Weight on earth (W) =?

Weight (W) = mass (m) × Acceleration due to gravity (g)

Weight (W) = 12 × 9.8

Weight on earth = 117.6 N

B. Weight on moon

Mass (m) = 12 KgAcceleration due to gravity on earth (g) = 9.8 m/s²  Acceleration due to gravity on moon (g) = (1/6) × 9.8 = 1.63 m/s² Weight on moon (W) =?

Weight (W) = mass (m) × Acceleration due to gravity (g)

Weight (W) = 12 × 1.63

Weight on moon = 19.56 N

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Complete question:

The force of gravity on the Moon is said to be one-sixth of that on the Earth. What would a mass of 12 kg weigh; (a) on the Earth (b) on the moon

A 900 pF capacitor is charged by 100 V battery. (a) How much electrostatic energy is stored by the capacitor? The capacitor is disconnected from the battery and connected to another 900 pF capacitor. How much is the electrostatic energy stored in the system?​

Answers

Answer:

The electrostatic energy stored in the system after connecting the two capacitors is 0.09 Joules

Explanation:

The formula to calculate the electrostatic energy stored in a capacitor is given by:

E = (1/2) * C * V^2

where:

E is the electrostatic energy,

C is the capacitance, and

V is the voltage across the capacitor.

(a) For the first scenario, where a 900 pF capacitor is charged by a 100 V battery:

C = 900 pF = 900 * 10^(-12) F

V = 100 V

Using the formula, we can calculate the electrostatic energy stored in the capacitor:

E = (1/2) * C * V^2

E = (1/2) * (900 * 10^(-12)) * (100^2)

E = 0.045 J

Therefore, the electrostatic energy stored by the capacitor in the first scenario is 0.045 Joules.

(b) In the second scenario, when the first capacitor is disconnected from the battery and connected to another 900 pF capacitor, the total capacitance in the system becomes:

C_total = C1 + C2

C_total = 900 pF + 900 pF

C_total = 1800 pF = 1800 * 10^(-12) F

The voltage across the capacitors remains the same, as they are connected in parallel.

Using the formula for electrostatic energy, we can calculate the new energy stored in the system:

E_total = (1/2) * C_total * V^2

E_total = (1/2) * (1800 * 10^(-12)) * (100^2)

E_total = 0.09 J

Therefore, the electrostatic energy stored in the system after connecting the two capacitors is 0.09 Joules.

(a) Define the following terms(1) internal energy (ii) an isovolumic process (b) state first law of thermodynamics (c
​

Answers

The definitions of Internal energy,  Isovolumic process and the law of thermodynamics are:

(a) (i) Internal energy: Internal energy refers to the total energy stored within a system, including the kinetic and potential energies of its particles. It is a state function and depends only on the current state of the system, such as its temperature, pressure, and volume.

(ii) Isovolumic process: An isovolumic or isochoric process is a thermodynamic process where the volume of the system remains constant. During an isovolumic process, no work is done by or on the system through expansion or compression.

(b) The first law of thermodynamics, also known as the law of conservation of energy, states that energy cannot be created or destroyed; it can only be transferred or converted from one form to another. The first law of thermodynamics is expressed mathematically as ΔU = Q - W, where ΔU is the change in internal energy of the system, Q is the heat transferred into or out of the system, and W is the work done on or by the system.

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