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3 years ago

5

For real gases, how does a change in pressure affect the ratio of pv to nrt?

2 answers:

Zielflug [23.3K]3 years ago

8 0

Explanation :

From real gas equation

PV = nRT

where,

P is the pressure of the gas

V is the volume of gas

n is the number of moles

R is the gas constant

T is the temperature

According to Gay-Lussac’s law, at constant volume pressure of the gas is directly proportional to the temperature on the Kelvin scale.

so, as pressure increases keeping volume constant then the temperature will also increase.

Hence, on increasing P the ratio of PV to nRT also increases.

bogdanovich [222]3 years ago

3 0

At constant temperature and volume the pressure of a gas is directly proportional to the number of moles of the gas. PV =n RT, where R is the universal gas constant. A change in pressure depends with the number of moles of the gas, such that if the number of moles increases then there are many vibrations and collision of the gas molecules with the walls of a container thus increasing the pressure and vice versa.

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Maksim231197 [3]

Answer:

v = 282.84 m / s

Explanation:

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.v = √ T /ρ

Where v is the speed of the wave, T the tension in the wire and ρ the density of the wire

If the tension is doubled

T = 2T₀

v = √ (2T₀ / ρ)

v = √2 √ T₀ / ρ

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calculate

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3 years ago

A car moves forward up a hill at 12 m/s with a uniform backward acceleration of 1.6 m/s2. What is its displacement after 6 s?

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Answer:

The displacement of the car after 6s is 43.2 m

Explanation:

Given;

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time of motion, t = 6 s

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3 years ago

The tub of a washing machine goes into its spin cycle, starting from rest and gaining angular speed steadily for 6.00 s, at whic

MatroZZZ [7]

Answer:

$N = 42 rev$

Explanation:

As we know that initial angular speed of the tub was zero and then it increases uniformly to 4 rev/s in t = 6.00 s

so we will have

$\theta_1 = \frac{\omega_f + \omega_i}{2} t$

$\theta_1 = \frac{2\pi f_2 + 2\pi f_1}{2}(t)$

$\theta_1 = \frac{2\pi \times 4 + 0}{2}(6)$

$\theta_1 = 75.4 rad$

Now when the tub will comes to rest uniformly after opening the lid in time interval of t = 15 s

then we have

$\theta_2 = \frac{\omega_f + \omega_i}{2} t$

$\theta_2 = \frac{2\pi f_2 + 2\pi f_1}{2}(t)$

$\theta_2 = \frac{2\pi \times 4 + 0}{2}(15)$

$\theta_2 = 188.5 rad$

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$\theta = \theta_1 + \theta_2$

$\theta = 75.4 + 188.5$

$\theta = 263.9 rad$

so number of revolutions is given as

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3 0

3 years ago