Which two elements are most abundant within the sun?

Order the follow processes from (1) the least work done by the system to (5) the most work done by one mole of an ideal gas at 2

quester [9]

Answer : The order of process from (1) the least work done by the system to (5) the most work done by the system will be:

(1) < (5) < (3) < (4) < (2)

Explanation :

<u>The formula used for isothermally irreversible expansion is :</u>

$w=-p_{ext}dV\\\\w=-p_{ext}(V_2-V_1)$

where,

w = work done

$p_{ext}$ = external pressure

$V_1$ = initial volume of gas

$V_2$ = final volume of gas

<u>The expression used for work done in reversible isothermal expansion will be,</u>

$w=-nRT\ln (\frac{V_2}{V_1})$

where,

w = work done = ?

n = number of moles of gas = 1 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas = $25^oC=273+25=298K$

$V_1$ = initial volume of gas

$V_2$ = final volume of gas

First we have to determine the work done for the following process.

(1) An isothermal expansion from 1 L to 10 L at an external pressure of 2.5 atm.

$w=-p_{ext}(V_2-V_1)$

$w=-(2.5atm)\times (10-1)L$

$w=-22.5L.atm=-22.5\times 101.3J=-2279.25J$

(2) A free isothermal expansion from 1 L to 100 L.

$w=-nRT\ln (\frac{V_2}{V_1})$

$w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{100L}{1L})$

$w=-11409.6J$

(3) A reversible isothermal expansion from 0.5 L to 4 L.

$w=-nRT\ln (\frac{V_2}{V_1})$

$w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{4L}{0.5L})$

$w=-5151.97J$

(4) A reversible isothermal expansion from 0.5 L to 40 L.

$w=-nRT\ln (\frac{V_2}{V_1})$

$w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{40L}{0.5L})$

$w=-10856.8J$

(5) An isothermal expansion from 1 L to 100 L at an external pressure of 0.5 atm.

$w=-p_{ext}(V_2-V_1)$

$w=-(0.5atm)\times (100-1)L$

$w=-49.5L.atm=-49.5\times 101.3J=-5014.35J$

Thus, the order of process from (1) the least work done by the system to (5) the most work done by the system will be:

(1) < (5) < (3) < (4) < (2)

8 0

3 years ago

Your boss tells you that water at 68 F is flowing in a cooling tower loop at a rate of 600 gpm, in an 8-inch inside diameter pip

avanturin [10]

Explanation:

The volumetric flow rate of water will be as follows.

q = $600 gpm \times \frac{0.000063 m^{3} s^{-1}}{1 gpm}$

= 0.0378 $m^{3}/s
$

Diameter = $8 in \times \frac{0.0254 m}{1 in}$

= 0.2032 m

Relation between area and diameter is as follows.

A = $\frac{\pi}{4} \times D^{2}$

= $\frac{3.14}{4} \times (0.2032 m)^{2}$

= 0.785 x 0.2032 x 0.2032

= 0.0324 $m^{2}$

Also, q = A × V

or, V = $\frac{q}{A}$

= $\frac{0.0378 m^{3}/s}{0.0324 m^{2}}$

= 1.166 m/s

As, viscosity of water = 1 cP = $10^{-3}$ Pa-s

Density of water = 1000 $kg/m^{3}$

Therefore, we will calculate Reynolds number as follows.

Reynolds number = $\frac{D \times V \times density}{viscosity}$

= $\frac{0.2032 m \times 1.166 m/s \times 1000}{10^{-3}}$

= 236931.2

Hence, the flow will be turbulent in nature.

Thus, we can conclude that the Reynolds number is 236931.2 and flow is turbulent.

8 0

3 years ago

If 200.4g of water is mixed with 101.42g of salt the mass of the final solution would be reported as

d1i1m1o1n [39]

Answer:

301.8 g

Explanation:

We prepare a solution with 200.4 g of water (solvent) and 101.42 g of salt (solute). The mass of the solution is equal to the sum of the mass of the solvent and the mass of the solute.

m(solution) = m(solute) + m(solvent)

m(solution) = 200.4 g + 101.42 g

m(solution) = 301.8 g (we round-off to one decimal according to the significant figures rules)

8 0

2 years ago

Match each of the following topical dosage forms with its correct definition.

Dmitry [639]

Answer:

cream - contains a higher proportion of oil than water

ointment - dr4g mixed in approximately equal proportions of oil and water

i don't know about the other two sorry

8 0

1 year ago

URGENT PLEASE HELP!!!

mote1985 [20]

Even though two grams seemed to disappear or vanish, the law of conversation of mass still holding true. Mercuric oxide, when heated, forms a gas of mercury and oxygen. During the investigation, some gas could have escaped or evaporated.

5 0

3 years ago

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