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masya89 [10]
3 years ago
14

Entropy change is evaluated using Eq. 6.2a based on an internally reversible process. Can the entropy change between two states

of a system be determined for any process linking the two states whether reversible or not?

Engineering
1 answer:
Vadim26 [7]3 years ago
4 0

Answer:

YES

Explanation:

Entropy is an extensive property of the system entropy change that value of entropy change can be determined for any process between the states whether reversible or not. i have attached the formula to calculate entropy change which is independent of whether the system is reversible or not and can be determined for any process.

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Calculate the osmotic pressure of seawater containing 3.5 wt % NaCl at 25 °C . If reverse osmosis is applied to treat seawater,
AlladinOne [14]

Answer:

Highest osmotic pressure that membrane may experience is

' =58.638 atm

Explanation:

Suppose sea-water taken is M= 1 kg

Density of water = 1000 kg/m3

Therefore Volume of water= Mass,M/Density of water

V= 1 kg/(1000 kg/m3)

V= 10-3 m3= 1 Litre

Since mass of Nacl is 3.5 wt%,Therefore in 1 kg of water

Mass present of NaCl= m= 0.035*1000 g

m= 35 g

Since molecular weight of NaCl= 58.44 g/mol =M.W.

Thus its Number of moles of Nacl= m/M.W

nNaCl= 35g/58.44 gmol-1

= 0.5989 mol

ans since volume of solution is 1 L thus concentration of NaCl is ,C= number of moles/Volume of solution in Litres

C= 0.5989mol/ 1L

=0.5989 M

Since 1 mol NaCL disssociates to form 2 moles of ions of Na+ andCl- Thus van't hoff factor i=2

And osmotic pressure  = iCRT ------------------------------(1)( Where R= 0.0821 L.atm/mol.K and T= 25oC= 298.15 K)

Putting in equation 1 ,we get  = 2*(0.5989 mol/L)*(0.0821 L.atm/mol.K)*298.15 K

=29.319 atm

Now as the water gets filtered out of the membrane,the water's volume decreases and concentration C of NacL increases, thus osmotic pressure also increases.Thus, at 50% water been already filtered out, the osmotic pressure at the membrane will be maximum

Thus Volume of water left after 50% is filtered out as fresh water= 0.5 L (assuming no salt passes through semi permeable membrane)

Thus New concentration of NaCl C'= 2*C

C'=2*0.5989 M

=1.1978 M

and Since Osmotic pressure is directly proportional to concentration, Thus As concentration C doubles to C', Osmotic Pressure  ' also doubles from  ,

Thus,Highest osmotic pressure that membrane may experience is,  '=2*  

=2*29.319 atm

' =58.638 atm

3 0
3 years ago
declare integer product declare integer number product = 0 do while product < 100 display ""Type your number"" input number p
Brilliant_brown [7]

Full Question

1. Correct the following code and

2. Convert the do while loop the following code to a while loop

declare integer product

declare integer number

product = 0

do while product < 100

display ""Type your number""

input number

product = number * 10

loop

display product

End While

Answer:

1. Code Correction

The errors in the code segment are:

a. The use of do while on line 4

You either use do or while product < 100

b. The use of double "" as open and end quotes for the string literal on line 5

c. The use of "loop" statement on line 7

The correction of the code segment is as follows:

declare integer product

declare integer number

product = 0

while product < 100

display "Type your number"

input number

product = number * 10

display product

End While

2. The same code segment using a do-while statement

declare integer product

declare integer number

product = 0

Do

display "Type your number"

input number

product = number * 10

display product

while product < 100

4 0
4 years ago
Steam enters an adiabatic turbine at 10 MPa and 500°C and leaves at 10 kPa with a quality of 90 percent. Neglecting the changes
Anna35 [415]

Answer:

The mass flow rate of steam m=5.4 Kg/s

Explanation:

Given:

  At the inlet of turbine P=10 MPa  ,T=500 C

 AT the exit of turbine  P=10 KPa   ,x=0.9

 Required power=5 MW

From steam table

<u> At 10 MPa and 500 C:</u>

  h=3374 KJ/Kg  ,s=6.59 KJ/Kg-K  (Super heated steam table)

<u>At 10 KPa:</u>

h_g=2675.1 KJ/Kg, h_f=417.51  KJ/Kg

s_g= 7.3  KJ/Kg-K                ,s_f=1.3   KJ/Kg-K

So enthalpy of steam at the exit of turbine

h= h_f+x(h_g- h_f)

Now by putting the values

h= 417.51+0.9(2675.1- 417.51) KJ/Kg

h=2449.34  KJ/Kg

Lets take m is the mass flow rate of steam

So 5\times 10^3=m\times (3374-2449.34)

m=5.4 Kg/s

So the mass flow rate of steam m=5.4 Kg/s

8 0
4 years ago
• Suppose that a particular algorithm has time complexity T(n) = 10 ∗ 2n, and that execution of the algorithm on a particular ma
elena-s [515]

Answer:

The number of inputs processed by the new machine is 64

Solution:

As per the question:

The time complexity is given by:

T(n) = 10\times 2n

where

n = number of inputs

T = Time taken by the machine for 'n' inputs

Also

The new machine is 65 times faster than the one currently in use.

Let us assume that the new machine takes the same time to solve k operations.

Then

T(k) = 64 T(n)

\frac{T(k)}{T(n)} = 64

\frac{20k}{20n} = 64

k = 64n

Thus the new machine will process 64 inputs in the time duration T

8 0
4 years ago
Who plays blox burg? if u do write ur username in commentsSsSS
marusya05 [52]
No thx but have a great day and god bless u :-)
8 0
3 years ago
Read 2 more answers
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