The signal propagation time between two ground stations in a synchronous satellite link is about millisecond is

For methyl chloride at 100°C the second and third virial coefficients are: B = −242.5 cm 3 ·mol −1 C = 25,200 cm 6 ·mol −2 Calcu

bogdanovich [222]

Answer:

a)W=12.62 kJ/mol

b)W=12.59 kJ/mol

Explanation:

At T = 100 °C the second and third virial coefficients are

B = -242.5 cm^3 mol^-1

C = 25200 cm^6 mo1^-2

Now according isothermal work of one mole methyl gas is

W=- $\int\limits^a_b {P} \, dV$

a= $v_2\\$

b= $v_1$

from virial equation

$\frac{PV}{RT}=z=1+\frac{B}{V}+\frac{C}{V^2}\\ \\P=RT(1+\frac{B}{V} +\frac{C}{V^2})\frac{1}{V}\\$

And

$W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V} } \, dV$

a= $v_2\\$

b= $v_1$

Now calculate V1 and V2 at given condition

$\frac{P1V1}{RT} = 1+\frac{B}{v_1} +\frac{C}{v_1^2}$

Substitute given values $P_1\\$ = 1 x 10^5 , T = 373.15 and given values of coefficients we get

$10^5(v_1)/8.314*373.15=1-242.5/v_1+25200/v_1^2$

Solve for V1 by iterative or alternative cubic equation solver we get

$v_1=30780 cm^3/mol$

Similarly solve for state 2 at P2 = 50 bar we get

$v_1=241.33 cm^3/mol$

Now

$W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V} } \, dV$

a=241.33

b=30780

After performing integration we get work done on the system is

W=12.62 kJ/mol

(b) for Z = 1 + B' P +C' P^2 = PV/RT by performing differential we get

dV=RT(-1/p^2+0+C')dP

Hence work done on the system is

$W=-\int\limits^a_b {P(RT(-1/p^2+0+C')} \, dP$

a= $v_2\\$

b= $v_1$

by substituting given limit and P = 1 bar , P2 = 50 bar and T = 373 K we get work

W=12.59 kJ/mol

The work by differ between a and b because the conversion of constant of virial coefficients are valid only for infinite series

8 0

3 years ago

Fill in the blank with the correct response.

antoniya [11.8K]

Answer:

Mentorship

Explanation:

Im smart.

5 0

3 years ago

Which of the following best characterizes a scientific law?

sesenic [268]

Answer:

subjective im pretty sure

Explanation:

6 0

3 years ago

In order to be a Mechanical Engineer, you need to:

djyliett [7]

Answer:

3

Explanation:

it is compulsory to have a bachelor's degree

4 0

3 years ago

Read 2 more answers

Air enters a compressor operating at steady state at 1.05 bar, 300 K, with a volumetric flow rate of 21 m3/min and exits at 12 b

omeli [17]

Answer:

- 46.5171kW

Explanation:

FIrst, the value given:

P1 = 1.05 bar (Initial pressure)

P2 = 12 bar (final pressure)

Heat transfer, Q = - 3.5 kW (It is negative because the compressor losses heat to the surroundings)

Mgaseous nitrogen = Mair = 28.0134 Kg/mol (constant)

Universal gas constant, Ru = 8.3143 Kj/Kgmolk

Specific gas constant, R = 0.28699 Kj/KgK

Initial temperature, T1 = 300 K

Final temperature, T2 = 400 K

Finding the volume:

P1V1 = RT1

V1 = RT1 ÷ P1

= (0.28699 Kj/KgK X 300k) ÷ 105

Note convert bar to Kj/Nm by multiply it by 100

V1 = 0.81997 m3/Kg

To get the mass flow rate:

m = volumetric flow rate / V1

= (21 m3/min x 1/60seconds) ÷ 0.81997 m3/Kg

= 0.4268Kg/s

Using tables for the enthalpy,

hT1 = 300.19 KJ/Kg

hT2 = 400.98 KJ/Kg

The enthalpy change = hT2 - hT1

= 100.79 KJ/Kg

Power, P = Q - (m X enthalpy change)

= - 3.5 - (0.4268 X 100.79)

= - 46.5171kW

3 0

3 years ago

The signal propagation time between two ground stations in a synchronous satellite link is about millisecond is​

The signal propagation time between two ground stations in a synchronous satellite link is about millisecond is