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kupik [55]
1 year ago
14

A system fitted with a piston expands when it absorbs 52.7J of heat from the surroundings. The piston is working against a press

ure of 0.693 atm. the final volume is 63.2L. what was the initial volume of the system if the internal energy of the system decreased by 107.4J?
Chemistry
1 answer:
Allushta [10]1 year ago
6 0

Answer: The initial volume of the system is 60.29 L.

Explanation:

According to the first law of thermodynamics,

      \Delta U = Q - W

As it is given that heat is being added to the system so, \Delta H will be positive. And, work done on the system is negative and work done by the system is positive.

So here, \Delta U = -107.4 J

                    Q = 52.7 J

                    P = 0.693 atm

And,     W = PdV

or,        W = P (V_{final} - V_{initial})

So,       \Delta U = Q - P (V_{final} - V_{initial})

            -107.4 J = 52.7 J - 0.693 \times 101.325 (63.2 - V)

            -160.1 = -43.79 - 70.21 (63.2 - V)

           63.2 - V = \frac{160.1 + 43.79}{70.21}

                  - V =  2.90 - 63.2

                     V = 60.29 L

Thus, we can conclude that the initial volume of the system is 60.29 L.

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Answer:

C₂H₇F₂P

Explanation:

Given parameters:

Composition by mass:

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                H = 7%

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                 P  = 31%

Unknown:

Empirical formula of compound;

Solution :

The empirical formula is the simplest formula of a compound. To solve for this, follow the process below;

                                   C                          H                         F                   P

% composition

by mass                     24                          7                        38                  31

Molar mass                 12                           1                         19                  31

Number of

moles                       24/12                          7/1                    38/19           31/31

                                     2                               7                       2                   1

Dividing

by the

smallest                      2/1                             7/1                       2/1                1/1

                                     2                                7                        2                   1

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Answer:

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