When sedimentary rock is exposed to heat and pressure, what does it change into?

Suppose that 4.4 moles of a monatomic ideal gas (atomic mass = 7.9 × 10-27 kg) are heated from 300 K to 500 K at a constant volu

timurjin [86]

Answer:

1) ΔQ₁ = 10.97 x 10³ J = 10.97 KJ

2) W₁ = 0 J

3) P = 41.66 x 10³ Pa = 41.66 KPa

4) v = 1618.72 m/s

5) ΔQ₂ = - 18.29 x 10³ J = - 18.29 KJ

6) W₂ = - 7.33 KJ

Explanation:

1)

The heat transfer for a constant volume process is given by the formula:

ΔQ₁ = ΔU = n Cv ΔT

where,

ΔQ₁ = Heat transfer during constant volume process

ΔU = Change in internal energy of gas

n = No. of moles = 4.4 mol

Cv = Molar Specific Heat at Constant Volume = 12.47 J/mol.k

ΔT = Change in Temperature = T₂ - T₁ = 500 k - 300 k = 200 k

Therefore,

ΔQ₁ = (4.4 mol)(12.47 J/mol.k)(200 k)

ΔQ₁ = 10.97 x 10³ J = 10.97 KJ

2)

Since, work done by gas is given as:

W₁ = PΔV

where,

ΔV = 0, due to constant volume

Therefore,

W₁ = 0 J

4)

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

K.E = (3/2)KT

but, K.E is also given by:

K.E = (1/2)mv²

Comparing both equations:

(1/2)mv² = (3/2)KT

mv² = 3KT

v = √(3KT/m)

where,

v = average speed of gas molecue = ?

K = Boltzman Constant = 1.38 x 10⁻²³ J/k

T = Absolute Temperature = 500 K

m = mass of a molecule = 7.9 x 10⁻²⁷ kg

Therefore,

v = √[(3)(1.38 x 10⁻²³ J/k)(500 k)/(7.9 x 10⁻²⁷ kg)]

v = 1618.72 m/s

3)

From kinetic molecular theory, we know that or an ideal gas:

P = (1/3)ρv²

where,

P = pressure of gas = ?

m = Mass of Gas = (Atomic Mass)(No. of Atoms)

m = (Atomic Mass)(Avogadro's Number)(No. of Moles)

m = (7.9 x 10⁻²⁷ kg/atom)(6.022 x 10²³ atoms/mol)(4.4 mol)

m = 0.021 kg

ρ = density = mass/volume = 0.021 kg/0.44 m³ = 0.0477 kg/m³

Therefore,

P = (1/3)(0.0477 kg/m³)(1618.72 m/s)²

P = 41.66 x 10³ Pa = 41.66 KPa

5)

The heat transfer for a constant pressure process is given by the formula:

ΔQ₂ = n Cp ΔT

where,

ΔQ₂ = Heat transfer during constant pressure process

n = No. of moles = 4.4 mol

Cp = Molar Specific Heat at Constant Pressure = 20.79 J/mol.k

ΔT = Change in Temperature = T₂ - T₁ = 300 k - 500 k = -200 k

Therefore,

ΔQ₂ = (4.4 mol)(20.79 J/mol.k)(-200 k)

ΔQ₂ = - 18.29 x 10³ J = - 18.29 KJ

Negative sign shows heat flows from system to surrounding.

6)

From Charles' Law, we know that:

V₁/T₁ = V₂/T₂

V₂ = (V₁)(T₂)/(T₁)

where,

V₁ = 0.44 m³

V₂ = ?

T₁ = 500 K

T₂ = 300 k

Therefore,

V₂ = (0.44 m³)(300 k)/(500 k)

V₂ = 0.264 m³

Therefore,

ΔV = V₂ - V₁ = 0.264 m³ - 0.44 m³ = - 0.176 m³

Hence, the work done , will be:

W₂ = PΔV = (41.66 KPa)(- 0.176 m³)

W₂ = - 7.33 KJ

Negative sign shows that the work is done by the gas

4 0

3 years ago

The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. It is a good approximation to assume e =

Svetllana [295]

Answer:

Explanation:

(a) Rigel, 2.7x10^32W, T = 11,000K

But L = 4pR²sT⁴

L = 2.7x10^32W, T = 11,000K, s= 5.67 x 10^-8, R= radius in meters

Rigel parallax, p = 0.00378 arc sec

Substituting the various values and making R the subject of the formula

R² = L/(4psT⁴)

R² = 2.7x10^32/(4 x 0.003878 x 5.67x10^-8 x (11,000)⁴)

R² = 2.7x10^32/1.2877x10^7

R² = 2.096761668 x 10^25

R = 4.579041021 x 10^12meters

(b)

Procyon B, 2.1x10^23W, T = 10,000K

But L = 4pR²sT⁴

L = 2.1x10^23W, T = 10,000K, s= 5.67 x 10^-8, R= radius in meters

Procyon B parallax, p = 0.00284 arc sec

R² = 2.1x10^23/(4 x 0.00284 x 5.67x10^-8 x (10,000)⁴)

R² = 2.1x10^23/(6.441 x 10^6)

R² = 3.26036 x 10^16

R = 1.80565 x 10^8 meters

(c) The radius of Rigel is given as 4.579041021 x 10^12meters and the radius of Procyon B is given by 1.80565 x 10^8 meters shows the remarkable difference between a super-giant star(Rigel) and a white dwarf star (Procyon B)

The radius of the sun a red star is 6.96 x 10^8meters which shows a certain level of resemblance with the size of a dwarf white star Procyon B.

The sun is larger than Procyon B as estimated above.

7 0

3 years ago

A coil is wrapped with 300 turns of wire on the perimeter of a circular frame (radius = 8.0 cm). Each turn has the same area, eq

TiliK225 [7]

Answer:

Approximately 18 volts when the magnetic field strength increases from $\rm 20\; mT$ to $\rm 80\;mT$ at a constant rate.

Explanation:

By the Faraday's Law of Induction, the EMF $\epsilon$ that a changing magnetic flux induces in a coil is:

$\displaystyle \epsilon = N \cdot \frac{d\phi}{dt}$ ,

where

$N$ is the number of turns in the coil, and
$\displaystyle \frac{d\phi}{dt}$ is the rate of change in magnetic flux through this coil.

However, for a coil the magnetic flux $\phi$ is equal to

$\phi = B \cdot A\cdot \cos{\theta}$ ,

where

$B$ is the magnetic field strength at the coil, and
$A\cdot \cos{\theta}$ is the area of the coil perpendicular to the magnetic field.

For this coil, the magnetic field is perpendicular to coil, so $\theta = 0$ and $A\cdot \cos{\theta} = A$ . The area of this circular coil is equal to $\pi\cdot r^{2} = \pi\times 8.0\times 10^{-2}\approx \rm 0.0201062\; m^{2}$ .

$A\cdot \cos{\theta} = A$ doesn't change, so the rate of change in the magnetic flux $\phi$ through the coil depends only on the rate of change in the magnetic field strength $B$ . The size of the magnetic field at the instant that $B = \rm 50\; mT$ will not matter as long as the rate of change in $B$ is constant.

$\displaystyle \begin{aligned} \frac{d\phi}{dt} &= \frac{\Delta B}{\Delta t}\times A \\&= \rm \frac{80\times 10^{-3}\; T- 20\times 10^{-3}\; T}{20\times 10^{-3}\; s}\times 0.0201062\;m^{2}\\&= \rm 0.0603186\; T\cdot m^{2}\cdot s^{-1}\end{aligned}$ .