One mole of an ideal gas is expanded from a volume of 1.00 liter to a volume of 6.43 liters against a constant external pressure of 1.00 atm. how much work (in joules) is performed on the surroundings? ignore significant figures for this problem. (t = 300 k; 1 l·atm = 101.3 j)

-550.059 J

Explanation:

To calculate the amount of work done for an isothermal process is given by the equation:

W = amount of work done = ?

P = pressure = 1.00 atm

= initial volume = 1.00 L

= final volume = 6.43 L

Putting values in above equation, we get:

To convert this into joules, we use the conversion factor:

So,

The negative sign indicates the system is doing work.

Hence, the work done on the surroundings is -550.059 J

"record the time to complete a chemical reaction" would most likely be part of a lab procedure.

the answer is option a.