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3 years ago

Calculate the hfg and sfg of steam at 120oC from the Clapeyron equation, and compare them to the tabulated values.

Chemistry

1 answer:

den301095 [7]3 years ago

5 0

Answer:

The tabulated values at 120 C of hfg = 2202.1 kJ/Kg and sfg = 5.6

kJ/Kg.K

Explanation:

According to Clapeyron Equation :

$h_{fg}=TV_{fg}\left (\frac{\partial P}{\partial T} \right )_{sat}$

$=T(V_{g}-V_{f})_{120^{0}}\left ( \frac{\Delta P}{\125^{0}-115^{0}} \right )$

Taking P2 at temperature 125 and P1 at 115 degree Celsius .

T = 120 + 273.15 = 393.15 K

P2 = 232.23 kPa and P1 = 169.18kPa

(these are experimental values given in the Data)

$= 393.15(0.88133 -0.001060\ m^{3}/Kg)\frac{232.23-169.18 kPa}{10}$

$h_{fg} = 2206.8 kJ/Kg$

$S_{fg}=\frac{h_{fg}}{T}$

$\frac{2206.8}{393.15}$

= 5.6131 kJ/Kg.K

The tabulated values at 120 C of hfg = 2202.1 kJ/Kg and sfg = 5.6131

kJ/Kg.K

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3 years ago

A horizontal cylinder equipped with a frictionless piston contains 785 cm3 of steam at 400 K and 125 kPa pressure. A total of 83

guapka [62]

Answer:

a. 478.69 K

b. 939.43 $cm^{3}$

c. 19.30 J

d. 64.5J

Explanation:

From the question, we can identify the following;

$V_{o}$ = 785 $cm^{3}$ = 0.000785 $m^{3}$

$T_{o}$ = 400K

$P_{o}$ = 125 Kpa = 125 000 Pa

Using the ideal gas equation,

PV = nRT

where R is the molar gas constant = 8.314 $m^{3}$ ⋅Pa⋅ $K^{-1}$ ⋅ $mol^{-1}$

Thus, n = PV/RT = (125000 × 0.000785)/(8.314 × 400) = 0.03 mol

a. Steam temperature in K

To calculate this, we use the constant pressure process;

q = nΔH

Where q is 83.8J according to the question

Thus;

83.8 = 0.03 × [34980 + 35.5 $T_{1}$ - (34980 + 35.5 $T_{o}$ )]

83.8 = (0.03 × 35.5) ( $T_{1}$ - 400K)

83.8 = 1.065 ( $T_{1}$ - 400)

78.69 = ( $T_{1}$ - 400)

$T_{1}$ = 400 + 78.69

$T_{1}$ = 478.69 K

b. Final cylinder volume

To calculate this, we make use of the Charles' law(Temperature and pressure are directly proportional)

$V_{1}$ / $T_{1}$ = $V_{o}$ / $T_{o}$

$V_{1}$ = $V_{o}$ $T_{1}$ / $T_{o}$

$V_{1}$ = (785 × 478.69)/400

$V_{1}$ = 939.43 $cm^{3}$

c. Work done by the system

Mathematically, the work done by the system is calculated as follows;

w = P( $V_{1}$ - $V_{o}$ ) = 125 KPa ( 939.43 - 785) = 19.30 J

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ΔU = q - w = 83.8 - 19.3 = 64.5J

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3 years ago

Imagine you had HCl with a concentration of exactly 0.10 mol/dm3. If 0.023 dm3 of a sodium hydroxide solution, NaOH (aq), could

polet [3.4K]

Answer:

Explanation:

Step 1: Calculate the amount of sodium hydroxide in moles

Volume of sodium hydroxide solution = 25.0 ÷ 1,000 = 0.0250 dm3

Rearrange:

Concentration in mol/dm3 =

Amount of solutein mol = concentration in mol/dm3 × volume in dm3

Amount of sodium hydroxide = 0.100 × 0.0250

= 0.00250 mol

Step 2: Find the amount of hydrochloric acid in moles

The balanced equation is: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

So the mole ratio NaOH:HCl is 1:1

Therefore 0.00250 mol of NaOH reacts with 0.00250 mol of HCl

Step 3: Calculate the concentration of hydrochloric acid in mol/dm3

Volume of hydrochloric acid = 20.00 ÷ 1000 = 0.0200 dm3

Concentration in mol/dm3 =

= 0.125 mol/dm3

Step 4: Calculate the concentration of hydrochloric acid in g/dm3

Relative formula mass of HCl = 1 + 35.5 = 36.5

Mass = relative formula mass × amount

Mass of HCl = 36.5 × 0.125

= 4.56 g

So concentration = 4.56 g/dm3

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3 years ago