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miskamm [114]
2 years ago
11

In order to fill a tank of 1000 liter volume to a pressure of 10 atm at 298K, an 11.5Kg of the gas is required. How many moles o

f the gas are present in the tank? What is the molecular weight of the gas? Assuming that the gas to be a pure element can you identify it?
Engineering
1 answer:
lesya [120]2 years ago
3 0

Answer:

The molecular weight will be "28.12 g/mol".

Explanation:

The given values are:

Pressure,

P = 10 atm

  = 10\times 101325 \ Pa

  = 1013250 \ Pa

Temperature,

T = 298 K

Mass,

m = 11.5 Kg

Volume,

V = 1000 r

   = 1 \ m^3

R = 8.3145 J/mol K

Now,

By using the ideal gas law, we get

⇒ PV=nRT

o,

⇒ n=\frac{PV}{RT}

By substituting the values, we get

       =\frac{1013250\times 1}{8.3145\times 298}

       =408.94 \ moles

As we know,

⇒ Moles(n)=\frac{Mass(m)}{Molecular \ weight(MW)}

or,

⇒        MW=\frac{m}{n}

                   =\frac{11.5}{408.94}

                   =0.02812 \ Kg/mol

                   =28.12 \ g/mol

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Which of the following is the BEST definition for e-commerce? *
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Answer:

(C) Buying and selling items electronically on the internet.

Explanation:

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2 years ago
The function below takes a single string parameter: input_string. If the input contains the lowercase letter z, return the strin
Whitepunk [10]

Answer:

# string_contains function is defined with input_string

# as arguments

def string_contains(input_string):

   # if-statement that check if letter z is in input_string

   if 'z' in input_string:

       # it print"has the letter z" if it has z

       print("has the letter z")

   else:

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# string_contains function is called

string_contains("The animal is zebra")        

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Explanation:

The code is written in Python and well commented.

Image of the output when the function is called is attached.

4 0
2 years ago
calculate how much black eyes seeds are necessary to plant a 6- hectare( 14.425 acres) field. given that the weight of 1000 blac
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Answer: 1.38g

Explanation:

Width of planting area = 100m

Field size = 6-hectares(14.425 acres)

Weight of 1000 black eye seed = 230g

1 lb = 453.4g

1 black eye seed = 230g/1000 = 0.23g = 0.00023kg

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6 hectare = 60,000sq metre

(Weight/Area) kg/m2

0.00023kg / 10,000 = 2.3×10^-8kg/m^2

And field size = 6hectares = 60,000m^2

(2.3×10^-8) × 60,000 = 0.00138kg of black eye seed

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6 0
3 years ago
Obtain a relation for the logarithmic mean temperature difference for use in the LMTD method?
kolezko [41]

Answer:

The log mean temperature difference is:

ΔT,lm=(ΔT1-ΔT2)/㏑(ΔT1/ΔT2)

Explanation:

To evaluate the equivalent average temperature difference between two fluids we consider a parallel-flow double-pipe heat exchanger (see attached diagram). The temperature of the hot and cold fluids is large at the inlet of the heat exchanger and decreases exponentially toward the outlet.  

We can assume that the outer surface of the heat exchanger is well insulated and that heat transfer only occurs between the two fluids. We can also assume negligible kinetic and potential. The energy balance on each fluid can be written as the rate of heat loss from the hot fluid is equal to the rate of heat gained by the cold fluid in any section of the heat exchanger:

Q = -m,h×c,ph×dT,h   (1)

where Q=rate of heat loss, m=mass flow rate, c,ph=heat capacity of the hot fluid, dT,h= differential temperature of the hot fluid

Q = m,c×c,pc×T.c  (2)

where Q=rate of heat loss, m=mass flow rate, c,ph=heat capacity of the cold fluid, dT,h= differential temperature of the cold fluid

The temperature of the hot fluid change is negative and is added to make Q positive. Solving equations 1 and 2 in terms of dT:

dT.h = - Q/(m,h×c,ph)

dT.c =  Q/(m,c×c,pc)

and taking the difference:

dT,h-dT,c= d(T,h - T,c) = -Q(1/(m,h×c,ph) + 1/(m,c×c,pc)) (3)

The heat transfer rate in the differential section of the heat exchanger can be expressed as:

Q = U(T,h-T,c)×dA,s  (4)

where U=overall heat transfer coefficients, dA,s = differential sectional area. Substitute equation 4 into 3:

d(T,h - T,c)/(T,h - T,c) = -U×dA,s×(1/(m,h×c,ph) + 1/(m,c×c,pc))  (5)

Integrating equation 5:

㏑((T,h out - T,c out)/(T,h in - T,c in)) = -U×A,s×(1/(m,h×c,ph) + 1/(m,c×c,pc))  (6)

The first law of thermodynamics requires the rate of heat transfer from hot and cold fluid to be equal.

Q= m×c, pc×(T, c out-T, c in)  (7)

Q= m×c, ph×(T,h out-T, h in)   (8)

Solve equations 7 and 8 for m,c×c, pc and m,h×c, ph and substituting into equation 6:

Q = U×A,s×ΔT,lm

Where the log mean temperature difference is:

ΔT,lm=(ΔT1-ΔT2)/㏑(ΔT1/ΔT2)

Download pdf
8 0
3 years ago
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