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3 years ago

In a system, when potential energy decreases, then entropy also decreases. True False

Physics

1 answer:

Paladinen [302]3 years ago

8 0

Answer: False

Explanation:

The potential energy is called organized energy because it is not related to the mass movement of the molecules.

Heat is disorganized energy because it is related to the mass movement of the molecules. Entropy is related to the randomness of the molecules.

If the randomness of the system increases then the entropy will increase. The entropy of the system either remains constant or increases.

The entropy of the system never decreases.

Therefore, the answer will be false.

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The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50∘C is 2000 m/s. Note that 1.0 mol of diatomic hydrogen at

denis-greek [22]

Answer:

A) d. (1/4)(2000m/s) = 500 m/s

B) c. 4000 J

C) f. None of the above (2149.24 m/s)

Explanation:

The translational kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 1^-23 J/K

T = Absolute Temperature

but,

K.E = (1/2) mv²

where,

v = root mean square velocity

m = mass of one mole of a gas

Comparing both equations:

(3/2)KT = (1/2) mv²

v = √(3KT)/m _____ eqn (1)

FOR HYDROGEN:

v = √(3KT)/m = 2000 m/s _____ eqn (2)

FOR OXYGEN:

velocity of oxygen = √(3KT)/(mass of oxygen)

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

using eqn (2)

velocity of oxygen = (1/4)(2000 m/s) = 500 m/s

K.E = (3/2)KT

Since, the temperature is constant for both gases and K is also a constant. Therefore, the K.E of both the gases will remain same.

K.E of Oxygen = K.E of Hydrogen

K.E of Oxygen = 4000 J

using eqn (2)

At, T = 50°C = 323 k

v = √(3KT)/m = 2000 m/s

m = 3(1.38^-23 J/k)(323 k)/(2000 m/s)²

m = 3.343 x 10^-27 kg

So, now for this value of m and T = 100°C = 373 k

v = √(3)(1.38^-23 J/k)(373 k)/(3.343 x 10^-27 kg)

v = 2149.24 m/s

8 0

3 years ago

The acceleration reaches its minimal value of zero at the top of the trajectory. *

Serhud [2]

it is true. the trajectory reaches the value of zero at the top

4 0

3 years ago

You and your friend are both 14 years old. Your friend however moves to a planet where just 30 minutes on that planet is the sam

vaieri [72.5K]

Answer:

ok so

Explanation:

1.) 1 hour

2.) 14 years old

3.)

24 years old

4.) you are

5 0

2 years ago

To lift a load of 100 kg a distance of 1 m an effort of 25 kg must be applied over an inclined plane of length 4 m. What must be

Troyanec [42]

Using conservation of energy law:-
∑ work in = ∑ work out
and work= force* displacement
so when we wanted to move a 100kg a distance of 1m
we multiplied 100*1 = work out
so work in should be equal to 100*g Joules, where g is the acceleration due to gravity.
so workout = 100*g = 25*g *x (divide both sides by 25*g)
x=4m

by the same way:-
------------------------
work in = 100kg * 2m * g (m/ $s^2$ )= work out
so work out = 25*x*g = 200* g (divide both sides by 25*g)
x=8m

6 0

3 years ago

Read 2 more answers

A 2 kg basketball has a momentum of 4 kg m/s. What is the ball's velocity?

Mumz [18]

That is the answer to the question
I hope this helps you.
Thank you for your question

4 0

2 years ago