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alukav5142 [94]
4 years ago
13

Pablo's experiments confirm the relationship between pressure and the number of gas molecules. as the pressure in a system incre

ases, the number of gas molecules increases. pablo conducted many trials. what can you conclude about the relationship between pressure and the number of gas molecules?
a.pressure and the number of gas molecules are not related.
b.pressure and the number of gas molecules remain constant.
c.pressure and the number of gas molecules are directly related.
d.pressure and the number of gas molecules are indirectly related.
Chemistry
2 answers:
horrorfan [7]4 years ago
6 0
The correct answer is:  [C]:  
___________________________________________________________
          "<span>pressure and the number of gas molecules are directly related."
___________________________________________________________
<u>Note</u>:  The conclusion was: "</span> as the pressure in a system increases, the number of gas molecules increases" — over the course of many trials.
          This means that the "pressure" and the "number of gas molecules" are directly related.

Furthermore, this conclusion is consistent with the "ideal gas law" equation:

     " PV = nRT " ; 
____________________________________________________________
in which:

    "P = Pressure" ; 
    "n = number of gas molecules" ; 
___________________________________________________________
        All other factors held equal, when "n" (the "number of gas molecules") 
 increases in value (on the "right-hand side" of the equation), the value for "P" (the "pressure" — on the "left-hand side" of the equation), increases.
___________________________________________________________
Tomtit [17]4 years ago
3 0

Answer is C .............

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Explanation:

The given data is as follows.

        m = 10.0 kg = 10,000 g    (as 1 kg = 1000 g)

      Initial temp. of block 1, T_{1} = 100^{o}C = (100 + 273) K = 373 K  

      Initial temp. of block 2, T_{2} = 0^{o}C = (0 + 273) K = 273 K

So, heat released by block 1 = heat gained by block 2

            mC \Delta T = mC \times \Delta T

  10000 g \times 0.385 \times (T_{f} - 100)^{o}C = 10000 g \times 0.385 \times (0 - T_{f})^{o}C

                  T_{f} - 100^{o}C = 0^{o}C - T_{f}    

                   2T_{f} = 100^{o}C

                          T_{f} = 50^{o}C

Convert temperature into kelvin as (50 + 273) K = 323 K.              

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                         = 1243550 J

or,                     = 1243.5 kJ

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            = 10000 g \times 0.385 J/K g \times -0.143

            = -554.12 J/K

Now, entropy change for block 2 is as follows.

   \Delta S_{2} = mC ln \frac{T_{f}}{T_{i}}

           = 10000 g \times 0.385 J/K g \times ln \frac{323}{273}

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            \Delta S_{total} = \Delta S_{1} + \Delta S_{2}

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