A brief note on Systems, Heat, Work and the First Law . Read it before a test. Try each question at the end before you open its answer.
Systems System: the part we study. Surroundings: everything else. The boundary separates them. Open: energy and matter cross. Closed: only energy. Isolated: neither. State variables: p p p , V V V , T T T and n n n . State functions ( p p p , V V V , T T T , U U U ) depend only on the state. Heat and work depend on the path. Extensive (mass, volume, U U U ) grow with the amount. Intensive ( T T T , p p p , density) do not. Heat, work and U U U U is the total energy stored in the system. We measure only Δ U \Delta U Δ U . Sign rule: energy into the system is positive. q > 0 q > 0 q > 0 for heat absorbed, w > 0 w > 0 w > 0 for work done on the system. Adiabatic: q = 0 q = 0 q = 0 , so Δ U = w \Delta U = w Δ U = w . The first law Δ U = q + w \Delta U = q + w Δ U = q + w . The energy of an isolated system is constant. Constant outside pressure: w = − p ex Δ V w = -p_\text{ex}\Delta V w = − p ex Δ V . 1 1 1 L bar = 100 = 100 = 100 J, 1 1 1 L atm = 101.3 = 101.3 = 101.3 J. Free expansion: w = 0 w = 0 w = 0 . Reversible, isothermal, ideal gas: w = − 2.303 n R T log V f V i w = -2.303\,nRT\log\dfrac{V_f}{V_i} w = − 2.303 n R T log V i V f , the most work possible. Isothermal ideal gas: Δ U = 0 \Delta U = 0 Δ U = 0 , so q = − w q = -w q = − w . A worked example A gas absorbs 500 J of heat and expands from 2.0 L to 5.0 L against a constant 1.2 bar. Find w w w and Δ U \Delta U Δ U .
Volume change. Δ V = 5.0 − 2.0 = 3.0 \Delta V = 5.0 - 2.0 = 3.0 Δ V = 5.0 − 2.0 = 3.0 L. Work. w = − 1.2 × 3.0 = − 3.6 w = -1.2 \times 3.0 = -3.6 w = − 1.2 × 3.0 = − 3.6 L bar = − 360 = -360 = − 360 J. First law. Δ U = 500 + ( − 360 ) = 140 \Delta U = 500 + (-360) = 140 Δ U = 500 + ( − 360 ) = 140 J. Practice with new numbers
A program to try The program works through the example above.
Run it. Compare each line with the worked example.
1. Change p_ext to 2.0. Does the internal energy still rise?
Mistakes to avoid Dropping the minus sign in w = − p ex Δ V w = -p_\text{ex}\Delta V w = − p ex Δ V . Leaving work in L bar when q q q is in J. Writing Δ q \Delta q Δ q or Δ w \Delta w Δ w . Heat and work are not state functions. Thinking q = 0 q = 0 q = 0 at constant temperature. For an ideal gas it is Δ U \Delta U Δ U that is zero. Check yourself Is a sealed can of soda in a fridge open, closed or isolated? Stuck? Show a hint Show the answerA gas gives out 30 J of heat and has 70 J of work done on it. What is Δ U \Delta U Δ U ? Stuck? Show a hint Show the answerA gas expands by 0.50 L against 4.0 bar. What is w w w ? Stuck? Show a hint Show the answerAn ideal gas expands into a vacuum. Why is w w w zero? Stuck? Show a hint Show the answer