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3 years ago

Suppose we have two identical boxes

Physics

1 answer:

Marianna [84]3 years ago

8 0

Answer:

Explanation:

In thermal contact, energy is exchanged between the two systems until thermal equilibrium is reached.

In thermal equilibrium, the temperature of system one is always equal to temperature of system two.

Also from second law of thermodynamic;

The entropy of an isolated system never decreases: in equilibrium, the entropy stays the same; otherwise the entropy increases until equilibrium is reached.

Since the two boxes are in thermal equilibrium, the entropy of system 1 is equal to the entropy of system 2.

However, a microstate is a specific microscopic configuration of a thermodynamic system that the system may occupy with a certain probability in the course of its thermal fluctuations.

Since system 2 = twice system 1

Then, microscopic configuration of system 2 is equal to twice microscopic configuration of system 1.

P₂ (Box(A+B))= P₁(2A), since the two boxes are identical.

Therefore, the number of microstates of system 2 is twice as high as those of system 1.

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The minute hand of a wall clock measures 16 cm from its tip to the axis about which it rotates. The magnitude and angle of the d

olya-2409 [2.1K]

Answer:

Explanation:

Given

Minute hand length =16 cm

Time at a quarter after the hour to half past i.e. 1 hr 45 min

Angle covered by minute hand in 1 hr is 360 and in 45 minutes 270

$|r|=\frac{3\times 2\pi r}{4}=75.408 cm$

$Angle =270^{\circ}$

(c)For the next half hour

Effectively it has covered 2 revolution and a quarter

$|r|=\frac{2\pi r}{4}=25.136 cm$

angle turned $=90^{\circ}$

(f)Hour after that

After an hour it again comes back to its original position thus displacement is same =25.136

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3 years ago

A 7.75-l flask contains 0.482 g of hydrogen gas and 4.98 g of oxygen gas at 65°c. What is the partial pressure of oxygen in the

dimulka [17.4K]

Answer:

0.558 atm

Explanation:

We must first consider that both gases behaves like ideal gases, so we can use the following formula: PV=nRT

Then, we should consider that, whithin a mixture of gases, the total pressure is the sum of the partial pressure of each gas:

P₀ = P₁ + P₂ + ....

P₀= total pressure

P₁=P₂= is the partial pressure of each gass

If we can consider that each gas is an ideal gas, then:

P₀= (nRT/V)₁ + (nRT/V)₂ +..

Considering the molecular mass of O₂:

M O₂= 32 g/mol

And also:

R= ideal gas constant= 0.082 Lt*atm/K*mol

T= 65°C=338 K

4.98 g O₂ = 0.156 moles O₂

V= 7.75 Lt

Then:

P°O₂=partial pressure of oxygen gas= (0.156x0.082x338)/7.75

P°O₂= 0.558 atm

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3 years ago

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</span>

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3 years ago

The bowling ball is whizzing down the bowling lane at 4 m/s. If the mass of the bowling ball is 7 kg, what is its kinetic energy

Lisa [10]

Kinetic Energy = 1/2xmassx(velocity)^2
Input values;
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3 years ago

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Answer:

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