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lara [203]
3 years ago
10

A sealed helium balloon with an internal pressure of 1.00 atm and a volume of 4.50 L at 293 K is moved. What volume will the bal

loon occupy at a place where the pressure is 0.600 atm and the temperature is 253 K?
Chemistry
1 answer:
bixtya [17]3 years ago
8 0

Answer:

6.48 L

Explanation:

From the question,

Applying

PV/T = P'V'/T'......................... Equation 1

P = initial pressure of the helium balloon, V =  Initial volume of the balloon, T = Initial temperature of the balloon, P' = Final pressure of the balloon, T' = Final temperature of the balloon, V' = Final volume of the balloon.

make V' the subject of the equation

V' = PVT'/P'T......................... Equation 2

Given: P = 1 atm, V = 4.5 L, T' = 253 K, T= 293 K, P' = 0.6 atm

Substitute these values into equation 2

V' = (4.5×1×253)/(0.6×293)

V' = 1138.5/175.8

V' = 6.48 L

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Explanation:

The given data is as follows.

          \lambda = 211 nm,          \sum = 6.17 \times 10^{3} mol/L/cm

           l = 1 cm,             7% < Transmittance < 85%

Suppose the aqueous solution follows Lambert-Beer's law. Therefore,

                   Absorbance = -log \frac{\text{Percentage transmittance}}{100}

Hence, for 7% transmittance the value of absorbance will be as follows.

                  Absorbance = -log \frac{7}{100}

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                 Absorbance = -log \frac{85}{100}

                           A_{2} = 0.07058

According to Lambert-Beer's law.

                  A = \sum \times l \times C

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                 \sum = molar extinction coefficient

                 C = concentration

Therefore, concentration for 7% absorbance is as follows.

                    A_{1} = \sum \times l \times C_{1}

                  C_{1} = \frac{1.155}{6.17 \times 10^{3} \times 1}

                                  = 0.187 \times 10^{-3} mol/L

                                  = 0.187 mmol/L

Concentration for 85% absorbance is as follows.

                   A_{2} = \sum \times l \times C_{2}

                  C_{2} = \frac{0.07058}{6.17 \times 10^{3} \times 1}

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                                  = 0.01144 mmol/L

Thus, we can conclude that linear range of phenol concentration is 0.01144 mmol/L to 0.187 mmol/L.

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