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Blababa [14]
2 years ago
4

Some hydrogen gas is enclosed within a chamber being held at 200∘C whose volume is 0.0250 m3. Initially, the pressure in the gas

is 1.50×106Pa (14.8 atm). The chamber is removed from the heat source and allowed to cool until the pressure in the gas falls to 0.950×106Pa. At what temperature T2 does this occur?
Physics
1 answer:
VARVARA [1.3K]2 years ago
4 0

The final temperature of the gas is 27^{\circ}C

Explanation:

Since the volume of the gas is constant (the volume of the chamber does not  change), we can apply the pressure's law, which states that:

"For a gas kept at constant volume, the pressure of the gas is proportional to its absolute temperature"

Mathematically:

\frac{p}{T}=const.

where

p is the gas pressure

T is its Kelvin temperature

For this problem, the equation can be written as

\frac{p_1}{T_1}=\frac{p_2}{T_2}

where we have:

p_1=1.50\cdot 10^6 Pa is the initial pressure

p_2=0.950\cdot 10^6 Pa is the final pressure

T_1=200^{\circ}C+273=473 K is the initial temperature

T_2 is the final temperature

And solving for T2,

T_2=\frac{p_2 T_1}{p_1}=\frac{(0.950\cdot 10^6)(473)}{1.50\cdot 10^6}=300 K

So the final temperature is

T_2 = 300 K - 273 = 27^{\circ}C

Learn more about ideal gases:

brainly.com/question/9321544

brainly.com/question/7316997

brainly.com/question/3658563

#LearnwithBrainly

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Answer:

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Explanation:

From the question we are told that

     The rate at which ATP molecules are used is R =  80 ATP/ s

       The energy provided by a single ATP is  E_{ATP} =  0.8 *  10^{-19} J

       The velocity of the kinesin is  v  =  800 nm/s =  800*10^{-9} m/s

The power provided by the ATP in one second is  mathematically represented as

       P =  E_{ATP}  *  R

substituting values

       P =  80 * 0.8*10^{-19 }

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Now  this power is mathematically represented as

       P  =  F *  v

Where  F  is  the force the kinesin is exerting

  Thus  

          F  =   \frac{P}{v}

substituting values

            F  =   \frac{6.4*0^{-18}}{800 *10^{-9}}

           F  =   8*10^{-12} \ N

7 0
2 years ago
If a drop is to be deflected a distance d = 0.350 mm by the time it reaches the end of the deflection plate, what magnitude of c
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Answer:

q = 4.87 X 10^ -14 C

Explanation:

As d=0.350 mm

The ink drop will be accelerated by the electric field between the plates:

a = F/m

d = a(D0 / v)^2 / 2 ...... 1

a = qE/m ............... 2

Substituting 2  into 1:

d = (qE/m)(D0 / v)^2 / 2

q = 2mdv^2 / [E(D0)^2]

q = 2(1.00e-11 kg)(3.50e-4 m)(15.0 m/s)^2 / [(7.70e4 N/C)(2.05e-2 m)^2]

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Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 20.0 m above water wit
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Answer:

Explanation:

Given that,

Height of the bridge is 20m

Initial before he throws the rock

The height is hi = 20 m

Then, final height hitting the water

hf = 0 m

Initial speed the rock is throw

Vi = 15m/s

The final speed at which the rock hits the water

Vf = 24.8 m/s

Using conservation of energy given by the question hint

Ki + Ui = Kf + Uf

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Ki is initial kinetic energy

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Then,

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A.

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irakobra [83]
QUESTION:-A beam of white light shines onto a sheet of white paper. An identical beam of light shines onto a mirror. The light is scattered from the paper and reflected from the mirror.
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