PMDC Unit 43 · 7 outcomes

Thermodynamics

Unit 7 is heat flow, work by expanding gases, the first law, and why C_p is bigger than C_v by R. Signs: on this site W is work done on the system unless a stem says ‘by’.

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7.1

Thermal equilibrium and heat

PMDC: Describe that thermal energy is transferred from a region of higher temperature to a region of lower temperature.

Heat is energy on the move because of a temperature difference. It flows hot → cold until both sides share one temperature (thermal equilibrium). It does not flow because of ‘more heat sitting there’ — temperature is the driver.

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heat:

T_{h o t}

→

T_{co l d}

heat flows $T_{h o t}$ → $T_{co l d}$ until T is the same

Hold these

Heat is in transit. Once it arrives, it is internal energy.

7.2

Specific heat vs molar specific heat

PMDC: Differentiate between specific heat and molar specific heat.

Specific heat c: energy to raise $1 kg$ by $1 K$ . Molar specific heat C: energy to raise 1 mole by $1 K$ . Same idea, different per-what. Gases need two molar heats: $C_{v}$ (volume locked) and $C_{p}$ (pressure locked).

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c_{m e t a l}

400

c_{w a t er} 4000 J

kg⁻¹ K⁻¹

$Q = m c Δ T (c : J k g^{- 1} K^{- 1})$
$Q = n C Δ T (C : J m o l^{- 1} K^{- 1})$

Hold these

c is per kilogram. C is per mole. Do not mix m and n.

7.3

Work in a volume change

PMDC: Calculate work done by a thermodynamic system during a volume change.

A gas pushes a piston out. It does work on the world. For a slow (quasi-static) change at pressure P, that work is P $Δ V$ . On a P–V diagram it is the area under the curve. Expanding: the system does work. Compressing: the world does work on the system.

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W_{b y} = P Δ V = 1 \times 1 0^{5} \times 2 \times 1 0^{- 3} = 200 J

$W_{b y} sy s t e m = P Δ V$ (constant P)
$W_{b y}$ = area under P–V curve
expansion ( $Δ V$ > 0): system does work

Hold these

Read the stem: work by or work on. They are opposite signs.

7.4

First law of thermodynamics

PMDC: Describe the first law in terms of the change in internal energy, the heating of the system, and work done on the system.

The first law is bookkeeping. Heat added to the system (Q) and work done on the system ( $W_{o n}$ ) both raise internal energy U.

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Δ U = 5 + (- 2) = 3 (W o n sy s t e m)

$Δ U = Q + W_{o n}$
if W is work by the system: $Δ U$ = Q − $W_{b y}$

Hold these

U is a state function. Q and W are not — they depend on the path.
Know which sign convention the question uses.

7.5

First law as conservation of energy

PMDC: Explain that the first law of thermodynamics expresses the conservation of energy.

Energy is not created in the cylinder. Heat and work are just two doors. What enters minus what leaves is the change in U. That is conservation of energy for a thermal system.

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Δ U = 5 + (- 2) = 3 (W o n sy s t e m)

energy in − energy $o u t = Δ U$

Hold these

A ‘perfect engine’ that dumps no heat would break the second law, not the first. First law still holds.

7.6

Molar specific heats of a gas

PMDC: Define specific heat and molar specific heats of a gas.

For a gas, holding volume fixed means all the heat goes into U (no expansion work). That heat per mole per kelvin is $C_{v}$ . Holding pressure fixed, the gas expands as well, so you must feed extra heat. That is $C_{p}$ . Always $C_{p}$ > $C_{v}$ .

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C v : Δ V = 0 C p : Δ P = 0

$Q_{v} = n C_{v} Δ T$
$Q_{p} = n C_{p} Δ T$
$C_{p} > C_{v}$

Hold these

Constant volume: no P $Δ V$ work. Constant pressure: there is P $Δ V$ work.

7.7

C_p − C_v = R

PMDC: Apply the first law to derive C_p − C_v = R for an ideal gas.

One mole, ideal gas, temperature up by $1 K$ . At constant volume: $Q_{v} = C_{v}, W_{b y}$ = 0, so $Δ U = C_{v}$ .

At constant pressure: $Q_{p} = C_{p}$ , the gas expands, $W_{b y}$ = P $Δ V$ = R (because $P V = R T, Δ T = 1 K$ ). $Δ U$ is the same (U of an ideal gas depends only on T), so $C_{p} = C_{v}$ + R.

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C p - C v = 8.3 C p = 29.1

ideal gas: $Δ U$ = n $C_{v} Δ T$ (any process)
$C_{p} - C_{v}$ = R (one mole)
$C_{p} - C_{v}$ = R (MDCAT: often per mole)

Hold these

The extra heat at constant pressure is the expansion work R $Δ T$ per mole.
$R = 8.31$ J mol⁻¹ K⁻¹. Some questions use 2 cal mol⁻¹ K⁻¹.

Unit 43 formula sheet

Write these until they are cheaper than thinking, then open the quiz.

heat flows $T_{h o t}$ → $T_{co l d}$ until T is the same
$Q = m c Δ T (c : J k g^{- 1} K^{- 1})$
$Q = n C Δ T (C : J m o l^{- 1} K^{- 1})$
$W_{b y} sy s t e m = P Δ V$ (constant P)
$W_{b y}$ = area under P–V curve
expansion ( $Δ V$ > 0): system does work
$Δ U = Q + W_{o n}$
if W is work by the system: $Δ U$ = Q − $W_{b y}$
energy in − energy $o u t = Δ U$
$Q_{v} = n C_{v} Δ T$
$Q_{p} = n C_{p} Δ T$
$C_{p} > C_{v}$
ideal gas: $Δ U$ = n $C_{v} Δ T$ (any process)
$C_{p} - C_{v}$ = R (one mole)
$C_{p} - C_{v}$ = R (MDCAT: often per mole)

Open the Unit 43 quiz