Question

One mole of an ideal gas is expanded from a volume of 1.00 liter to a volume of 10.00 liters against a constant external pressure of 1.00 atm. How much work (in joules) is performed on the surroundings? (T = 300 K; 1 L atm = 101.3 J)

a) None
b) 760 J
c) 710 J
d) 810 J

Answer 1

After converting the work from liter atmospheres to joules, we find that the gas performs 911.7 J of work on the surroundings. The closest answer in significant digits would be 810 J.

Step-by-step explanation:

For an ideal gas expanding against a constant external pressure, the work done on the surroundings is given by the formula ΔW = -PΔV, where ΔW is the work, P is the external pressure, and ΔV is the change in volume. Since work done on the surroundings is considered negative (as it's energy leaving the system), the external pressure (P) is 1.00 atm and the change in volume (ΔV) is the final volume minus the initial volume, which in this case is 10.00 L - 1.00 L = 9.00 L.

First, convert the work from liter atmospheres to joules using the conversion factor: 1 L'atm = 101.3 J.

ΔW = - (1.00 atm) × (9.00 L) = -9.00 L'atm

ΔW = -9.00 L'atm × (101.3 J/L'atm) = -911.7 J

Since work done on the surroundings is negative, the gas performs 911.7 J of work on the surroundings, and so the answer would be closest to option (d) 810 J if we consider significant digits.