A machine employs the isothermal expansion of 4.2 moles of ideal gas from 3.72 Lto 9.68 L. At 293°C, the machine performs 4.49 kJ of work. What percent of the maximum possible work is the machine producing? (Do not include units in your answer. If you round during your calculation make sure to keep at least 3 decimal places. Report your answer to two decimal places.) Answer: 3.30 X The correct answer is: 44.45

44.455% 

Note this answer as wll as those of the substeps are rounded off to 3 decimal places.

Step-by-step explanation

Given: n = 4.2 moles

  T = 29.3°C

  V₂ = 9.68 L

  V₁ = 3.72 L

  W_actual = 4.49 kJ

Required: % work

Concept needed

(1) W_max = nRTln(V1?V2??)

(2) % work = Wmax?Wactual??×100

where W_max -- maximum/theoretical work 

  n -- moles

  R -- universal gas constant

  T -- absolute temperature

  V₂ -- final volume

  V₁ -- initial volume

  W_actual -- actual work

Additional Data

R = 8.314 x 10^-3 kJ/mol.K

Conversion

T, K = T,°C + 273.15

Solution

Step 1: Convert the temperature to kelvin

T = 29.3 + 273.15

   = 302.45 K

Step 2: Calculate the maximum work

W_max = (4.2 moles)(8.314 x 10^-3mol.KkJ?)(302.45 K)(ln 3.72L9.68L?)

 = 10.100 kJ

Step 3: Calculate % work

% work = 10.100kJ4.49kJ?×100

    = 44.455%