Final answer:
The work done by a gas that expands from 3 dm^3 to 13 L against a constant external pressure of 1 atm is 10 atm dm^3 (10 L-atm). We find this by using the formula W = -PΔV and taking the negative sign to indicate work done on the gas.
Step-by-step explanation:
The question involves calculating the work done by a gas during an expansion at constant external pressure, a concept from Chemistry. To find the work done, we use the formula:
W = -PΔV
Where:
- W is the work done by the gas (in liter-atmospheres, L-atm)
- P is the constant external pressure (in atmospheres, atm)
- ΔV is the change in volume (in liters, L)
The initial volume (Vi) = 3 dm3 (which is equivalent to 3 L because 1 dm3 = 1 L), final volume (Vf) = 13 L, and constant external pressure (P) = 1 atm. We substitute these values into the formula:
W = -1 atm × (13 L - 3 L)
W = -1 atm × 10 L
W = -10 L-atm
Since work done by the gas is work done against the external pressure, we take the negative sign into account; this indicates work done on the gas. Thus, the work done by the gas during the process is 10 atm dm3 or 10 L-atm.