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GrogVix [38]
2 years ago
12

What is the resulting pressure when you transfer gas from cylinder with a temperature of 22°C and the pressure of four atm into

a container with a temperature of 65°C? Assume volume remains constant.
Chemistry
1 answer:
Hatshy [7]2 years ago
4 0
Answer:
             P₂  =  4.58 atm

Solution:
               According to <span>Gay-Lussac's law ," the pressure of a given mass of an ideal gas at constant volume is directly proportional to absolute temperature".
For initial and final states of a gas, the relation is given as,

                                           P</span>₁ / T₁  =  P₂ / T₂   ----- (1)

Data Given;
                   P₁  =  4 atm

                   T₁  =  22 °C + 273  =  295 K

                   P₂  =  ??

                   T₂  =  65 °C + 273  =  338 K

Now, solving equation 1 for P₂,

                  P₂  =  P₁ T₂ / T₁

Putting values,

                  P₂  =  (4 atm × 338 K) ÷ 295 K

                  P₂  =  4.58 atm
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How much energy must be removed from a 125 g sample of benzene (molar mass= 78.11 g/mol) at 425.0 K to liquify the sample and lo
riadik2000 [5.3K]

Answer : The energy removed must be, -67.7 kJ

Solution :

The process involved in this problem are :

(1):C_6H_6(g)(425.0K)\rightarrow C_6H_6(g)(353.0K)\\\\(2):C_6H_6(g)(353.0K)\rightarrow C_6H_6(l)(353.0K)\\\\(3):C_6H_6(l)(353.0K)\rightarrow C_6H_6(l)(335.0K)

The expression used will be:

\Delta H=[m\times c_{p,g}\times (T_{final}-T_{initial})]+m\times \Delta H_{vap}+[m\times c_{p,l}\times (T_{final}-T_{initial})]

where,

\Delta H = heat released by the reaction = ?

m = mass of benzene = 125 g

c_{p,g} = specific heat of gaseous benzene = 1.06J/g^oC

c_{p,l} = specific heat of liquid benzene = 1.73J/g^oC

\Delta H_{vap} = enthalpy change for vaporization = 33.9kJ/mole=33900J/mole=\frac{33900J/mole}{78.11g/mole}J/g=434.0J/g

Molar mass of benzene = 78.11 g/mole

Now put all the given values in the above expression, we get:

\Delta H=[125g\times 1.06J/g.K\times (353.0-(425.0))K]+125g\times -434.0J/g+[125g\times 1.73J/g.K\times (335.0-353.0)K]

\Delta H=-67682.5J=-67.7kJ

Therefore, the energy removed must be, -67.7 kJ

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The correct response is the second option.
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