Question

The vaporization of 1 mole of liquid water (the system) at 100.9°C, 1.00 atm, is endothermic.

H2​O(l)+40.7kJ H2​O(g)

Assume that at exactly 100.0°C and 1.00 atm total pressure, 1.00 mole of liquid water and 1.00 mole of water vapor occupy 18.80 mL and 30.62 L, respectively.

Calculate the work done on or by the system when 1.25 mol of liquid H₂O vaporizes. (answer in joules)

I got -3874.9 J but the homework system keeps saying it is wrong.

Answer 1

The answer as per my calculations also yielded the same answer. So the answer must be correct. Make sure your answer input is as per the significant figures demanded by the system.

My calculations are as follows:

1. Change in volume = (1.25 mol) x (30.62 - 0.01880) L
2. Hence, Work done = -Pressure x change in volume.
3. W = -(101325 Pa ) x ((1.25 mol) x (30.62 - 0.01880) x 10^-3m³) = -3875.8 J

Also, the work done by the system = -3875.8 J

But work done on the system = + 3875.8 J

So try changing sign and make sure you use correct significant figures.

(If this all doesn't work please let me know via comments)