5) A sample of liquid butan-1-ol, CH,OH was brought to the boil in an open calor

Question

5) A sample of liquid butan-1-ol, CH,OH was brought to the boil in an open calorimeter. An electric current of 289 mA from a 12.0 V source was then passed through a resistive heater coil which was immersed in the liquid. The current was allowed to flow for a period of 245 s, during which time the temperature remained constant and 1.416 g of butan-1-ol was found to have evaporated. Calculate the molar enthalpy of vaporization of butan-1-ol 6) In the calibration step of a thermochemistry experiment, a current of 117 mA, from a 24.0 V source was allowed to flow through the electrical heater for 247 s and was found to result in an increase in the temperature of the calorimeter and its contents of 1.25 K. Calculate the heat capacity of the calorimeter and its contents. 7) The constant pressure molar heat capacity of methane, CH is 35.31 JK mol at temperatures close to 298 K. Calculate the enthalpy change when 2.00 mol of methane is heated from a temperature of 278 K to 318 K. 8) When 229 J of energy is supplied as heat to 3.00 mol Ar(g), the temperature of the sample increases by 2.55 K. Calculate the molar heat capacity at constant volume and constant pressure of the gas. 9) When 3.0 mol O,(g) is heated at a constant pressure of 3.25 atm, its temperature increases from 260 K to 285 K. Given the molar heat capacity of O, at constant pressure is 29.4 JK mol', calculate q, H and OU 10) Calculate 1, w, DU and H when 50 grams of O (g) is compressed isothermally and reversibly from 1 bar to 20 bar at 50°C.

Explanation / Answer

7. Given : Cp (molar hear capacity at constant pressure = 35.31 JK-1 mol-1

                  n = number of moles = 2 mol , T1 = 278 K and T2 = 318 K

                H = ?

Solution : according to 1st law of thermodynamics

Cp = H/T = H/(T2 – T1)

On rearranging the equation and substituting the given values

H = Cp (T2 – T1)

H = 35.31 JK-1 mol-1 X (318 -278)K

H = 1412.4 J mol-1    ( K get cancels) This is for one mole

For 2 mole

Enthalpy change for 2 mole of sample = 1412.4 J mol-1X 2 mol = 2824.8 J

8. Given : Cv (molar hear capacity at constant volume) = ?            JK-1 mol-1

                  n = number of moles = 3 mol Ar(g)

               T = T2 – T1 = 2.55 K

            q = heat supplied = 299 J

Solution : according to 1st law of thermodynamics

U=q+w

At constant volume work done w=0

Then U=q=229J

heat capacity at constant volume Cv = U/T = 299 J / 2.55K = 89.80 JK-1

Molar heat capacity at constant volume Cv = 89.80 JK-1 / 3 mol

Cv = 29.93 JK-1 mol-1

We know that Cp-Cv = R gas constant (R=8.314 J K-1 mol-1)

Above we got Cv = 29.93 JK-1 mol-1

Then Cp = R + Cv = (8.314 + 29.93) JK-1 mol-1

Molar heat capacity at constant pressure Cp= 38.244 JK-1 mol-1

9. Given : Cp (molar hear capacity at constant pressure) = 29.4 JK-1 mol-1

                  n = number of moles = 3 mol O2(g), T1 = 260 K and T2 = 285 K

               T = T2 – T1 = 285-260 = 25 K

                P = 3.25 atm

Solution : according to 1st law of thermodynamics

Molar heat capacity at constant pressure Cp = H/T

On rearranging the equation and substituting the given values

H = Cp T

H = 29.4 JK-1 mol-1 X 25K

H = 735 J mol-1    ( K get cancels) For one mole

For 3 moles

Enthalpy change for 2 mole of sample = 1412.4 J mol-1X 3 mol = 2205 J

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