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Viktor [21]
3 years ago
11

Vapor pressure is related to the temperature of the liquid. user: in an open system, the vapor pressure is equal to the _____. i

nside air pressure outside air pressure inside temperature outside temperature
Physics
2 answers:
Naddik [55]3 years ago
8 0
In an open system the vapor pressure of a liquid, if it is not boliling is less thatn the atmospheric pressure.

As the temperature of the liquid increases, the vapor pressure increases. When the vapor pressure equals the atmospheric pressure the molecules of liquid escape and the liquid boils. The temperature at which this happens is the boiling point.
iVinArrow [24]3 years ago
6 0

Answer

-Directly;  outside air pressure

Vapor pressure is directly related to the temperature of the liquid. user: in an open system, the vapor pressure is equal to the outside air pressure.

Explanation;

-As the temperature of a system increases, the average kinetic energy of the molecules increases in both the liquid and gas phases.

-A higher average kinetic energy facilitates the escape of molecules from the liquid phase into the gas phase. At the same time, the rate of return of gas phase molecules to the liquid also increases. A new equilibrium point is reached at a higher gaseous vapor pressure. The increase in vapor pressure with temperature is exponential.


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Nresistors, each having resistance equal to 1 2, are arranged in a circuit first in series and then in parallel. What is the rat
aleksandr82 [10.1K]

Answer:

option (b)

Explanation:

Let the resistance of each resistor is R.

In series combination,

The effective resistance is Rs.

rs = r + R + R + .... + n times = NR

Let V be the source of potential difference.

Power in series

Ps = v^2 / Rs = V^2 / NR ..... (1)

In parallel combination

the effective resistance is Rp

1 / Rp = 1 / R + 1 / R + .... + N times

1 / Rp = N / R

Rp = R / N

Power is parallel

Rp = v^2 / Rp = N V^2 / R    ..... (2)

Divide equation (1) by equation (2) we get

Ps / Pp = 1 / N^2

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3 years ago
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3 0
3 years ago
Read 2 more answers
A system gains 1500 J of heat, while the internal energy of the system increases by 4500 J and the volume decreases by . Assume
Assoli18 [71]

Answer:

Hence the pressure is 3\times 10^5 Pa

Explanation:

Given data

Q=1500 J   system gains heat

ΔV=- 0.010 m^3     there is a decrease in volume

ΔU= 4500 J        internal energy decrease

We know work done is

W= Q- ΔU

=1500-4500= -3000 J

The change in the volume at constant pressure is

ΔV= W/P

there fore P = W/ΔV= -3000/-0.01= 3×10^5

Hence the pressure is 3\times 10^5 Pa

3 0
3 years ago
A certain heat engine takes in 300 J of energy from a hot source and then transfers 200 J of that energy to a colder object. Wha
Greeley [361]

Answer:

The efficiency is 0.33, or 33%.

Explanation:

From the thermodynamics equations, we know that the formula for the efficiency of a heat engine is:

\eta=1-\frac{Q_2}{Q_1}

Where η is the efficiency of the engine, Q_1 is the heat energy taken from the hot source and Q_2 is the heat energy given to the cold object. So, plugging the given values in the formula, we obtain:

\eta=1-\frac{200J}{300J}=0.33

This means that the efficiency of the heat engine is 0.33, or 33% (The efficiency of an engine is dimensionless).

5 0
3 years ago
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