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3 years ago

What happens to the pressure of a gas inside a container if the temperature of the gas decreases?

Physics

2 answers:

serg [7]3 years ago

6 0

Answer:

Explanation:

According to the ideal gas equation

P V = n RT

where, P is pressure, V is volume , n is the number of moles, R is the gas constant and T is the temperature of the gas.

According to the equation, the pressure of the gas is directly proportional to the temperature of the gas. As the temperature decreases, the pressure of the gas also decreases.

My name is Ann [436]3 years ago

5 0

Answer:

Pressure Increases

Explanation:

When the temperature of a gas is increased inside a container then the pressure of the gas inside the container increases.

This can be deduced by the Ideal Gas law equation:

$P.V=n.R.T$

where:

P = pressure of the gas

V = volume of the gas

n = no. of moles of gas

R = universal gas constant

$T=$ temperature of the gas

So,

$P\propto T$ hence the pressure of the gas will increase while the temperature is increased at a constant volume

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A Smart Car, which has a mass of 1000 kg, is going 20 m/s. When it hits the barrier, it stops with a time of 0.5 seconds. What i

antiseptic1488 [7]

Answer:

The change in momentum = -20000 kg m/s.

Explanation:

Mass m = 1000 kg

speed v₁ = 20 m/s

speed v₂ = 0 m/s

We know that,

The change in momentum

ΔP = m (Δv)

ΔP = m (v₂ - v₁)

= 1000 (0 - 20)

= 1000 (-20)

= -20000 kg m/s

Thus, the change in momentum = -20000 kg m/s.

Note: negative sign indicates that the velocity is reducing when it hits the barrier.

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2 years ago

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I would say that it would take her 35 * 2 cashing Bill properly because I multiply 0.25 times 16 which gave me 1.50 + 2.50 equals 3.50

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3 years ago

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In 1993 Ileana Salvador of Italy walked 3.0km in under 12min. Suppose that during her walk Salvador is observed to steadily incr

padilas [110]

The distance covered is 115 m

Explanation:

The motion of Ileana is a uniformly accelerated motion (constant acceleration), therefore we can use the following suvat equation:

$s=(\frac{u+v}{2})t$

where

s is the distance covered

u is the initiaal velocity

v is the final velocity

t is the time elapsed

In this problem, we have:

u = 4.20 m/s

v = 5.00 m/s

t = 25.0 s

Therefore, we can re-arrange the equation to find the distance covered:

$s=(\frac{4.20+5.00}{2})(25.0)=115 m$

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3 years ago

At an accident scene on a level road, investigators measure a car’s skid mark to be 88 m long. The accident occurred on a rainy

Oduvanchick [21]

Answer:

The the speed of the car is 26.91 m/s.

Explanation:

Given that,

distance d = 88 m

Kinetic friction = 0.42

We need to calculate the the speed of the car

Using the work-energy principle

work done = change in kinetic energy

$W=\Delta K.E$

$\mu\ mg\times d=\dfrac{1}{2}mv^2$

$v^2=2\mu g d$

Put the value into the formula

$v=\sqrt{2\times0.42\times9.8\times88}$

$v=26.91\ m/s$

Hence, The the speed of the car is 26.91 m/s.

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3 years ago

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I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (wi

Masteriza [31]

Answer:

After it has traveled through 1 cm : $I(1 \ cm) = 0.5488 I_0$

After it has traveled through 2 cm : $I(2 \ cm) = 0.3012 I_0$

After it has traveled through 1 cm : $od( 1\ cm) = 0.2606$

After it has traveled through 2 cm : $od( 2\ cm) = 0.5211$

Explanation:

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient $\mu$ the formula is:

$I(x) = I_0 e^{-\mu x}$

where I is the intensity of the beam, $I_0$ is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

$I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = 0.5488 \ I_0$

After travelling 2 cm:

$I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}$

$I(2 \ cm) = I_0 e^{- 1.2}$

$I(2 \ cm) = 0.3012 \ I_0$

The optical density od is given by:

$od(x) = - log_{10} ( \frac{I(x)}{I_0} )$ .

So, after travelling 1 cm:

$od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )$

$od( 1\ cm) = - log_{10} ( 0.5488 )$

$od( 1\ cm) = - ( - 0.2606)$

$od( 1\ cm) = 0.2606$

After travelling 2 cm:

$od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )$

$od( 2\ cm) = - log_{10} ( 0.3012 )$

$od( 2\ cm) = - ( - 0.5211)$

$od( 2\ cm) = 0.5211$

3 0

3 years ago