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Sati [7]
3 years ago
6

A buret is a device designed to precisely dispense liquids. The scale is calibrated in mL with the zero point at the top with nu

mbers increasing down the buret. Assuming the buret was filled to 0.00 mL, determine the liquid volume dispensed by reading the buret pictured below, taking care to record the proper number of significant figures consistent with the design of the buret.
Engineering
1 answer:
kobusy [5.1K]3 years ago
6 0

Answer:

As there was no attached picture, I will explain how to take the measurement of liquids in any buret which you can then apply to the specific question

Explanation:

A buret is a  laboratory apparatus used to precisely measure the volume of liquids (usually alkalise or bases) used in a titration experiment. The standard buret has a capacity of 50 ml  and graduated in 0.1ml though burets with smaller capacities exist.

From the question, your buret is filled to the top (0.00ml) with liquid. It is very important when taking buret readings to place the buret below your eye level so that the bottom meniscus (lower part of the liquid) can be read.

To take the buret reading, note your initial buret reading (in this case 0.00ml) then titrate the liquid base in the buret against the acid by opening the tap located at the bottom of the buret.

When the titration or reaction is complete, note the final reading against the calibration of buret. You can do this by observing the lower meniscus of the liquid remaining in the buret. (Remember to keep the buret at eye level  to avoid parallax error),

The difference between your final buret reading and the initial buret reading gives you the precise volume of liquid used in the reaction.

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For methyl chloride at 100°C the second and third virial coefficients are: B = −242.5 cm 3 ·mol −1 C = 25,200 cm 6 ·mol −2 Calcu
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Answer:

a)W=12.62 kJ/mol

b)W=12.59 kJ/mol

Explanation:

At T = 100 °C the second and third virial coefficients are

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Now according isothermal work of one mole methyl gas is

W=-\int\limits^a_b {P} \, dV

a=v_2\\

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from virial equation  

\frac{PV}{RT}=z=1+\frac{B}{V}+\frac{C}{V^2}\\   \\P=RT(1+\frac{B}{V} +\frac{C}{V^2})\frac{1}{V}\\

And  

W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V}  } \, dV

a=v_2\\

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Now calculate V1 and V2 at given condition

\frac{P1V1}{RT} = 1+\frac{B}{v_1} +\frac{C}{v_1^2}

Substitute given values P_1\\ = 1 x 10^5 , T = 373.15 and given values of coefficients we get  

10^5(v_1)/8.314*373.15=1-242.5/v_1+25200/v_1^2

Solve for V1 by iterative or alternative cubic equation solver we get

v_1=30780 cm^3/mol

Similarly solve for state 2 at P2 = 50 bar we get  

v_1=241.33 cm^3/mol

Now  

W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V}  } \, dV

a=241.33

b=30780

After performing integration we get work done on the system is  

W=12.62 kJ/mol

(b) for Z = 1 + B' P +C' P^2 = PV/RT by performing differential we get  

         dV=RT(-1/p^2+0+C')dP

Hence work done on the system is  

W=-\int\limits^a_b {P(RT(-1/p^2+0+C')} \, dP

a=v_2\\

b=v_1

by substituting given limit and P = 1 bar , P2 = 50 bar and T = 373 K we get work  

W=12.59 kJ/mol

The work by differ between a and b because the conversion of constant of virial coefficients are valid only for infinite series  

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