Question

5) 0.022 kg of CO2 is compressed isothermally and reversibly at 298 K from initial pressure of 100 kPa
when the work obtained is 1200 J, calculate the final pressure. (Ans=263.4kPa)
6) Define the following terms:-
(1) Enthalpy of vaporization (ii) Standard enthalpy of combustion.
Why work done in vacuum is zero?
Long answer questions (4 Marks)​

Answer 1

Answer:

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Answer 2

Answer:

5)263.4KPA

Explanation:

GIVEN: MASS OF CO2=0.022kg=22g

             Temperature = T = 298 K

              Initial pressure = P1 = 100 kPa

              Work done on system = Wmax = 1200 J

To find: Final pressure = P2

Formula: Wmax = – 2.303 nRT log10 PI÷P2

Solution: Number of moles of CO2 = n = 22÷44=0.5 mol

1200 J = − 2.303 × 0.5 mol × 8.314 J K–1 mol–1 × 298 K × log10 100÷p2

log10×100÷p2=-1200÷2.303×0.5×8.314×298

∴p2=100÷0.3796=263.4kpa