Engineer's Notebook §1.2

First law

Energy is conserved. Heat Q added to a closed system either raises its internal energy U or leaves as work W done by the system:

ΔU = Q − W
U
internal energy
J
Q
heat added to the system
J
W
work done by the system
J
n
amount of gas
mol
cv
molar heat capacity, constant volume
J/(mol·K)
cp
molar heat capacity, constant pressure
J/(mol·K)
R
molar gas constant, 8.314
J/(mol·K)
T
temperature
K
p
pressure
Pa
V
volume
m³
Two bars. The top bar is the heat added. The bottom bar splits the same length into the rise in internal energy and the work done by the gas. HEAT IN, Q Q = 5 kJ GOES TO ΔU = 3.57 kJ W = 1.43 kJ
Fig. 1.2-1Q = ΔU + W: the two lower parts always add up to the heat in.
ΔT172 K
W1.43 kJ
ΔU = Q − W3.57 kJ

Worked example

Uses the values in Fig. 1.2-1. Change them and the working updates.

Given
  • Qheat added5 kJ
  • namount of gas1 mol
  • cvdiatomic gas (air, N₂, O₂)2.5R
  • processconstant pressure
Find

The change in internal energy ΔU.

Solution
c = cₚ = cᵥ + R = 29.1 J/(mol·K) ΔT = Q/nc = 5,000 J / (1 mol × 29.1 J/(mol·K)) = 172 K W = nRΔT = 1 mol × 8.314 J/(mol·K) × 172 K = 1.43 kJ ΔU = Q − W = 5 kJ − 1.43 kJ = 3.57 kJ
Answer
ΔU = 3.57 kJ

71.4% of the heat went into internal energy, raising the temperature by 172 K.

Related cases

  • Isothermal, ideal gas

    ΔU = 0Q = W = nRT ln(V2V1)

  • Adiabatic, no heat

    Q = 0ΔU = −W

  • Complete cycle

    ΔU = 0Qnet = Wnet

Assumptions
  • The system is closed: no mass crosses its boundary.
  • Heat into the system and work done by it are positive. Some texts count work done on the system instead and write ΔU = Q + W.
  • The gas is ideal with constant heat capacities, so U depends on temperature alone.
  • The process is slow enough that the gas is always close to equilibrium.
  • Changes in the system's bulk kinetic and potential energy are ignored.

Derivation
  1. Energy is conserved, so heat in either stays in the gas or leaves as work: Q = ΔU + W.
  2. For an ideal gas, U depends only on temperature: ΔU = ncvΔT.
  3. At constant pressure the gas pushes its boundary out: W = p ΔV = nR ΔT.
  4. So Q = n(cv + R)ΔT = ncpΔT, which gives ΔT from Q.

At constant volume no work is done, so all of the heat raises U: Q = ncvΔT.