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Brilliant_brown [7]
3 years ago
13

2.1. Three moles of an ideal gas (with temperature-independent CP = (7/2)R, CV = (5/2)R) is contained in a horizontal piston/cyl

inder arrangement. The piston has an area of 0.1 m2 and mass of 500 g. The initial pressure in the piston is 101 kPa. Determine the heat that must be extracted to cool the gas from 375°C to 275°C at: (a) constant pressure; (b) constant volume.
Chemistry
1 answer:
Alenkasestr [34]3 years ago
6 0

Answer:

The answers to the question are

The heat that must be extracted to cool the gas from 375°C to 275°C at

(a) Constant pressure Q = -6110.78 J

(b) Constant volume Q = -4365 J

Explanation:

Cp = (7/2)R

CV = (5/2)R)

Area of piston = 0.1 m²

Mass of piston, m = 500 g =0.5 kg

Mass of piston = 500 g

Initial pressure of piston p₁ = 101 kPa

Initial temperature of piston = 375 °C = 648.15‬ K

Final temperature of piston =275 °C  = ‪‪548.15‬ K

Pressure from the piston = m×g/0.1 =0.5×9.81/0.1 = 49.05 Pa

Pressure from gas alone = 101000-49.05= 100950.95 Pa

Therefore we have from the general gas equation

P₁V₁ = n·R·T₁ Where

V₁ = Initial volume piston = n·R·T₁/P₁  R = 8.314 J / mol·K.

V₁ = (8.314×2.1×648.15)/100950.95 = 0.112097  m³

For V₂ =n·R·T₂/P₂ = (8.314×2.1×548.15)/101000 = 0.094756 m³

\frac{V_1}{V_2} = \frac{T_1}{T_2} V₂ = T₂/T₁×V₁ = 548.15/648.15×0.11204  = 0.09476 m³

But p·V = m·R·T ⇒ m·R = P·V/T =100950.95*0.112097/648.15 = 17.45938

For constant pressure we have Q = m*cp*(T₂-T₁) =7/2·R*m*(T₂-T₁) =

7/2*17.45938*(548.15 -648.15) = -6110.7838 J =

(b) At constant volume, the heat that must be extracted =

We have P₁/T₁ = P₂/T₂

m*cv*(T₂-T₁)

= 5/2·R*m*(548.15-648.15) =5/2*17.46*(-100) = -4365 J

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Complete Question

You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution was 7.82 when exactly 13 mL of base had been added, you notice that the concentration of the unknown acid was 0.1 M. What is the pKa of your unknown acid?

Answer:

The pK_a value is pK_a  =7.82

Explanation:

From the question we are told

    The volume of base is  V_B = 26.mL = 0.0260L

     The pH of solution is  pH =  7.82

      The concentration of the acid is C_A = 0.1M

From the pH we can see that the titration is between a strong base and a weak acid

 Let assume that the the volume of acid is  V_A = 18mL= 0.018L

Generally the concentration of base

                    C_B = \frac{C_AV_A}{C_B}

Substituting value  

                     C_B = \frac{0.1 * 0.01800}{0.0260}

                    C_B= 0.0692M

When 13mL of the base is added a buffer is formed

The chemical equation of the reaction is

           HA_{(aq)} + OH^-_{(aq)} --------> A^{+}_{(aq)} + H_2 O_{(l)}

Now before the reaction the number of mole of base is  

            No \ of \ moles[N_B]  =  C_B * V_B

Substituting value  

                    N_B = 0.01300 * 0.0692

                         = 0.0009 \ moles    

                                 

Now before the reaction the number of mole of acid is  

            No \ of \ moles  =  C_B * V_B

Substituting value  

                    N_A = 0.01800 *0.1

                         = 0.001800 \ moles

Now after the reaction the number of moles of  base is  zero  i.e has been used up

    this mathematically represented as

                         N_B ' = N_B - N_B = 0

    The  number of moles of acid is  

             N_A ' = N_A  - N_B

                   = 0.0009\ moles

The pH of this reaction can be mathematically represented as

                 pH  = pK_a + log \frac{[base]}{[acid]}

Substituting values

                  7.82 = pK_a +log \frac{0.0009}{0.0009}

                  pK_a  =7.82        

                     

             

                                 

       

           

                     

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