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strojnjashka [21]
3 years ago
13

Measurements show that the enthalpy of a mixture of gaseous reactants decreases by during a certain chemical reaction, which is

carried out at a constant pressure. Furthermore, by carefully monitoring the volume change it is determined that of work is done on the mixture during the reaction. Calculate the change in energy of the gas mixture during the reaction. Be sure your answer has the correct number of significant digits. Is the reaction exothermic or endothermic?
Physics
1 answer:
snow_tiger [21]3 years ago
5 0

The given question is incomplete. The complete question is as follows.

Measurements show that the enthalpy of a mixture of gaseous reactants decreases by 338 kJ during a certain chemical reaction, which is carried out at a constant pressure. Furthermore, by carefully monitoring the volume change it is determined that 187 kJ of work is done on the mixture during the reaction. Calculate the change in energy of the gas mixture during the reaction. Be sure your answer has the correct number of significant digits. Is the reaction exothermic or endothermic ?

Explanation:

The given data is as follows.

    Change in enthalpy (\Delta H) = -338 kJ (as it is a decrease)

    Work done = 187 kJ,

    Change in energy (\Delta E) = ?

Now, according to the first law of thermodynamics the formula is as follows.

          \Delta H = \Delta E + P \Delta V

Hence, putting the given values into the above formula as follows.

        \Delta H = \Delta E + P \Delta V

Also, we know that W = P \Delta V

so,           \Delta H = \Delta E + W      

              -338 kJ = \Delta E + 187

              \Delta E = -151 kJ

Thus, we can conclude that the change in energy of the gas mixture during the reaction is -151 kJ.

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Explanation:

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v_{0_{1}}: is the initial speed of ball 1 = 0 (it is dropped from rest)

y = h - \frac{1}{2}gt^{2}     (1)

Now, for ball 2 we have:

y = y_{0_{2}} + v_{0_{2}}t - \frac{1}{2}gt^{2}    

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y_{0_{2}}: is the initial height of ball 2 = 0

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By equating equation (1) and (2) we have:

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v_{f_{1}} = - gt = -g*\sqrt{\frac{h}{2g}} = - 0.71\sqrt{gh}

The minus sign is because ball 1 is going down.

For ball 2:

v_{f_{2}} = v_{0_{2}} - gt = \sqrt{2gh} - g*\sqrt{\frac{h}{2g}} = (\sqrt{1} - \frac{1}{\sqrt{2}})*\sqrt{gh} = 0.41\sqrt{gh}

Therefore, taking the magnitude of ball 1 we can see that it has a greater speed than ball 2 when they are passing.

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For ball 1:

y_{1} = h - \frac{1}{2}gt^{2} = h - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h

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y_{2} = v_{0_{2}}t - \frac{1}{2}gt^{2} = \sqrt{2gh}*\sqrt{\frac{h}{2g}} - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h

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I hope it helps you!                  

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