1.what PROCESS describes the water moving through a membrane?

Air is compressed from an inlet condition of 100 kPa, 300 K to an exit pressure of 1000 kPa by an internally reversible compress

ElenaW [278]

Answer:

(a) $W_{isoentropic}=8.125\frac{kJ}{mol}$

(b) $W_{polytropic}=7.579\frac{kJ}{mol}$

(c) $W_{isothermal}=5.743\frac{kJ}{mol}$

Explanation:

Hello,

(a) In this case, since entropy remains unchanged, the constant $k$ should be computed for air as an ideal gas by:

$\frac{R}{Cp_{air}}=1-\frac{1}{k} \\\\\frac{8.314}{29.11} =1-\frac{1}{k}\\$

$0.2856=1-\frac{1}{k}\\\\k=1.4$

Next, we compute the final temperature:

$T_2=T_1(\frac{p_2}{p_1} )^{1-1/k}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.4}=579.21K$

Thus, the work is computed by:

$W_{isoentropic}=\frac{kR(T_2-T_1)}{k-1} =\frac{1.4*8.314\frac{J}{mol*K}(579.21K-300K)}{1.4-1}\\\\W_{isoentropic}=8.125\frac{kJ}{mol}$

(b) In this case, since $n$ is given, we compute the final temperature as well:

$T_2=T_1(\frac{p_2}{p_1} )^{1-1/n}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.3}=510.38K$

And the isentropic work:

$W_{polytropic}=\frac{nR(T_2-T_1)}{n-1} =\frac{1.3*8.314\frac{J}{mol*K}(510.38-300K)}{1.3-1}\\\\W_{polytropic}=7.579\frac{kJ}{mol}$

(c) Finally, for isothermal, final temperature is not required as it could be computed as:

$W_{isothermal}=RTln(\frac{p_2}{p_1} )=8.314\frac{J}{mol*K}*300K*ln(\frac{1000kPa}{100kPa} ) \\\\W_{isothermal}=5.743\frac{kJ}{mol}$

Regards.

8 0

3 years ago

On a cold morning, the smoke coming out of a chimney is observed to be in blue color. What could be the reason? Name the effect

nirvana33 [79]

Stack effect? I'm not totally sure about this...

Explanation:

Large amount of tiny particles of water droplets, dust and smoke are present on a misty day. These tiny particles in the air scatter blue colour of white light passing through it. When this scattered light reaches our eyes, the smoke appears blue.

6 0

3 years ago

The density of liquid mercury is 13.69 g/cm^3. How many atoms of mercury are in a 15.0 cm^3 sample? Use "E" for "x10" and use si

balu736 [363]

The formula is m = D x V
D = 13.69 g/cm^3.
V = 15.0 cm^3
the mass of the liquid mercury is m= 13.69 g/cm^3 x 15.0 cm^3 = 195g
the molar mass of Hg is 200,
1 mole of Hg = 200g Hg, so #mole of Hg= 195 / 200 = 0.97 mol
but we know that
1 mole = 6.022 E23 atoms
0.97 mole=?

6.022 E23 atoms x 0.97 / 1 mole = 5.84 E23 atoms

6 0

3 years ago

A sample of the chiral molecule limonene is 62% enantiopure. what percentage of each enantiomer is present? what is the percent

ikadub [295]

The mixture contains 62 % one isomer and 38 % the enantiomer.

Let’s say that the mixture contains 62 % of the (R)-isomer.

Then % (S) = 100 % -62 % = 38 %

ee = % (R) - % (S) = 62 % -38 % = 24 %

6 0

3 years ago

Read 2 more answers

each body cell of a goldfish contains 94 chromosomes.how many chromosomes are contained in a goldfish sex cell?

Shalnov [3]

47 I believe not 100% sure

8 0

2 years ago