Chemistry Problems: Internal Energy, Heat, and Work

Key Concepts

• Internal Energy (ΔU): Change in internal energy of a system.
• Heat (Q): Energy absorbed or released by the system.
  • Q is positive for endothermic processes (heat absorbed).
  • Q is negative for exothermic processes (heat released).
• Work (W): Work done on or by the system.
  • W is positive when work is done on the system.
  • W is negative when work is done by the system.
• First Law of Thermodynamics: ΔU = Q + W (in Chemistry)

Problem 1

• Given: 300 J of heat absorbed; 400 J of work done on system.
• Solution:
  • Q = +300 J (heat absorbed).
  • W = +400 J (work done on).
  • ΔU = Q + W = 300 J + 400 J = 700 J.
• Visualization:
  • System gains 700 J, surroundings lose 700 J.
  • Endothermic for system, exothermic for surroundings.

Problem 2

• Given: 700 J of heat released; 300 J of work done by system.
• Solution:
  • Q = -700 J (heat released).
  • W = -300 J (work done by).
  • ΔU = Q + W = (-700 J) + (-300 J) = -1000 J.
• Visualization:
  • System loses 1000 J, surroundings gain 1000 J.

Problem 3

• Given: Surroundings gain 250 J of heat; 470 J of work performed by surroundings.
• Solution:
  • Q = -250 J (heat flows to surroundings).
  • W = +470 J (work on system by surroundings).
  • ΔU = Q + W = (-250 J) + 470 J = 220 J.

Problem 4

• Given: Surroundings release 300 J of heat; 550 J of work done by system.
• Solution:
  • Q = +300 J (heat absorbed by system).
  • W = -550 J (work done by system).
  • ΔU = Q + W = 300 J - 550 J = -250 J.

Problem 5

• Task: Work done by gas expanding from 25 L to 40 L against 2.5 atm.
• Solution:
  • Formula: W = -PΔV.
  • ΔV = 40 L - 25 L = 15 L.
  • W = - (2.5 atm)(15 L) = -37.5 L·atm.
  • Convert to Joules: 1 L·atm = 101.3 J.
  • W = -37.5 L·atm × 101.3 J/L·atm = -3799 J.

Problem 6

• Task: Work required to compress gas from 50 L to 35 L at 8 atm.
• Solution:
  • ΔV = 35 L - 50 L = -15 L.
  • W = -(8 atm)(-15 L) = 120 L·atm.
  • Convert to Joules: W = 120 L·atm × 101.3 J/L·atm = 12156 J.

Problem 7

• Given: 500 J absorbed, expansion from 30 L to 70 L at 2.8 atm.
• Solution:
  • ΔV = 70 L - 30 L = 40 L.
  • W = -(2.8 atm)(40 L) = -112 L·atm.
  • Convert to Joules: W = -112 L·atm × 101.3 J/L·atm = -11345.6 J.
  • ΔU = 500 J + (-11345.6 J) = -10845.6 J.