Question

A system expands from a volume of 1.00 l to 2.00 l against a constant external pressure of 1.00 atm. What is the work (w) done by the system?

1) 3.00 j
2) 1.01 �� 102 j
3) 1.00 j
4) 1.01 �� 105 j
5) 2.00 j

Answer 1

The work done by the system during the expansion against a constant external pressure of 1.00 atm, when it expands from 1.00 L to 2.00 L, is 1.01 x 10^2 joules or 101.3 J.

Step-by-step explanation:

To calculate the work done by the system during an expansion against a constant external pressure, we use the formula w = -PΔV, where 'w' is the work done on the gas (in joules), 'P' is the constant external pressure (in atm) and 'ΔV' is the change in volume (in liters). Because work done by the system is considered negative in thermodynamics as the system loses energy, the formula includes a negative sign. However, when we calculate work done by the system (as asked in the question), we can ignore the negative sign because we are looking for the magnitude of work.

First, we need to convert the pressure from atm to the appropriate unit (joules per liter). We have that 1 atm = 101.3 J/L. Then we calculate the change in volume, which is 2.00 L - 1.00 L = 1.00 L. So, the work done by the system is 1.00 atm × 1.00 L = 1.00 L·atm. Converting this to joules, we get 1.00 L·atm × 101.3 J/L·atm = 101.3 J.

Therefore, the correct answer is 1.01 × 102 J or option 2).