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Valentin [98]
3 years ago
5

The contents of the cylinder is then entirely transfered to a partially filled holding tank of volume 10,000 L originally at pre

sure 2 atm and 300 K. What is the new pressure in the holding tank at 300 K?
Physics
1 answer:
Dahasolnce [82]3 years ago
7 0

Answer:

The new pressure in the holding tank is 2 atm.

Explanation:

Given that,

Volume = 10000 L

Initial pressure 2 atm

Initial temperature = 300 k

We need to calculate the new pressure

Using equation of ideal gas

PV=RT

R=\dfrac{PV}{T}

Where,P = pressure

T = temperature

V = volume

R = constant

Put the value into the formula

Volume of gas are remain constant

So,

\dfrac{P_{i}V}{T_{i}}=\dfrac{P_{f}V}{T_{f}}

\dfrac{2\times10000}{300}=\dfrac{P_{f}\times10000}{300}

P_{f}=2\ atm

Hence, The new pressure in the holding tank is 2 atm.

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140-160

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200-250

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2 years ago
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Why does a lunar eclipse last for a few hours whereas a solar eclipse only lasts for a few minutes?
Aneli [31]

Answer:

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8 0
2 years ago
A rock is lifted by a machine to a height of 10m. If it has a mass of 22 kilograms
zmey [24]

Answer:

2156J

Explanation:

Given parameters:

Height of lift  = 10m

Mass  = 22kg

Unknown:

Work done by the machine  = ?

Solution:

Work done is the force applied to move a body through a certain distance.

So;

        Work done  = Force x distance

Here;

       Work done  = mass x acceleration due to gravity x height

      Work done  = 22 x 9.8 x 10  = 2156J

5 0
3 years ago
Calculat the acceleration of a person at latitude<br> 40degreesowing to the rotation of the earth.
beks73 [17]

Answer:

acceleration of person = 9.77 m/s²

Explanation:

given data

latitude = 40 degree

to find out

Calculate the acceleration of a person

solution

we know that here 40 degree = 0.698 rad

so

acceleration of person = g - ω²R    ...............1

and 1 rotation complete in 24 hours = 360 degree

here g is 9.81

so we know Earth angular speed ω = 7.27 × 10^{-5} rad/s and R is earth radius that is 6.37 × 10^{6} m

so

put here value in equation 1 we get

acceleration of person = g - ω²R

acceleration of person = 9.81 - (7.27 × 10^{-5})² × 6.37 × 10^{6}

acceleration of person = 9.77 m/s²

6 0
3 years ago
A sample of monatomic ideal gas occupies 5.00 L at atmospheric pressure and 300 K (point A). It is warmed at constant volume to
leonid [27]

Answer:

(a) 0.203 moles

(b) 900 K

(c) 900 K

(d) 15 L

(e) A → B, W = 0, Q = Eint = 1,518.91596 J

B → C, W = Q ≈ 1668.69974 J Eint = 0 J

C → A, Q = -2,531.5266 J, W = -1,013.25 J, Eint = -1,518.91596 J

(g) ∑Q = 656.089 J, ∑W =  655.449 J, ∑Eint = 0 J

Explanation:

At point A

The volume of the gas, V₁ = 5.00 L

The pressure of the gas, P₁ = 1 atm

The temperature of the gas, T₁ = 300 K

At point B

The volume of the gas, V₂ = V₁ = 5.00 L

The pressure of the gas, P₂ = 3.00 atm

The temperature of the gas, T₂ = Not given

At point C

The volume of the gas, V₃ = Not given

The pressure of the gas, P₃ = 1 atm

The temperature of the gas, T₂ = T₃ = 300 K

(a) The ideal gas equation is given as follows;

P·V = n·R·T

Where;

P = The pressure of the gas

V = The volume of the gas

n = The number of moles present

R = The universal gas constant = 0.08205 L·atm·mol⁻¹·K⁻¹

n = PV/(R·T)

∴ The number of moles, n = 1 × 5/(0.08205 × 300) ≈ 0.203 moles

The number of moles in the sample, n ≈ 0.203 moles

(b) The process from points A to B is a constant volume process, therefore, we have, by Gay-Lussac's law;

P₁/T₁ = P₂/T₂

∴ T₂ = P₂·T₁/P₁

From which we get;

T₂ = 3.0 atm. × 300 K/(1.00 atm.) = 900 K

The temperature at point B, T₂ = 900 K

(c) The process from points B to C is a constant temperature process, therefore, T₃ = T₂ = 900 K

(d) For a constant temperature process, according to Boyle's law, we have;

P₂·V₂ = P₃·V₃

V₃ = P₂·V₂/P₃

∴ V₃ = 3.00 atm. × 5.00 L/(1.00 atm.) = 15 L

The volume at point C, V₃ = 15 L

(e) The process A → B, which is a constant volume process, can be carried out in a vessel with a fixed volume

The process B → C, which is a constant temperature process, can be carried out in an insulated adjustable vessel

The process C → A, which is a constant pressure process, can be carried out in an adjustable vessel with a fixed amount of force applied to the piston

(f) For A → B, W = 0,

Q = Eint = n·cv·(T₂ - T₁)

Cv for monoatomic gas = 3/2·R

∴ Q = 0.203 moles × 3/2×0.08205 L·atm·mol⁻¹·K⁻¹×(900 K - 300 K) = 1,518.91596 J

Q = Eint = 1,518.91596 J

For B → C, we have a constant temperature process

Q = n·R·T₂·㏑(V₃/V₂)

∴ Q = 0.203 moles × 0.08205 L·atm/(mol·K) × 900 K × ln(15 L/5.00 L) ≈ 1668.69974 J

Eint = 0

Q = W ≈ 1668.69974 J

For C → A, we have a constant pressure process

Q = n·Cp·(T₁ - T₃)

∴ Q = 0.203 moles × (5/2) × 0.08205 L·atm/(mol·K) × (300 K - 900 K) = -2,531.5266 J

Q = -2,531.5266 J

W = P·(V₂ - V₁)

∴ W = 1.00 atm × (5.00 L - 15.00 L) = -1,013.25 J

W = -1,013.25 J

Eint = n·Cv·(T₁ - T₃)

Eint = 0.203 moles × (3/2) × 0.08205 L·atm/(mol·K) × (300 K - 900 K) = -1,518.91596 J

Eint = -1,518.91596 J

(g) ∑Q = 1,518.91596 J + 1668.69974 J - 2,531.5266 J = 656.089 J

∑W = 0 + 1668.69974 J -1,013.25 J = 655.449 J

∑Eint = 1,518.91596 J + 0 -1,518.91596 J = 0 J

5 0
3 years ago
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