Heat capacities Cp, Cv; relationship Cp − Cv = nR (ideal gas): Exercises

Quick reminders (each is a plain-word restatement, not new notation):

$C_{v}$ = heat per mole per kelvin at fixed volume.
$C_{p}$ = heat per mole per kelvin at fixed pressure.
Mayer: $C_{p} - C_{v} = R$ (per mole), or $C_{p} - C_{v} = n R$ for $n$ moles.
$γ = C_{p} / C_{v}$ (the "ratio", used in adiabatic work).
For an ideal gas, $Δ U = n C_{v} Δ T$ always, and $Δ H = n C_{p} Δ T$ always.

L1 — Recognition

Exercise 1

A monatomic ideal gas has $C_{v} = \frac{3}{2} R$ . State $C_{p}$ and $γ$ .

Recall Solution (L1.1)

WHAT we do: apply Mayer's relation directly — it just says "add $R$ to $C_{v}$ ". $C_{p} = C_{v} + R = \frac{3}{2} R + R = \frac{5}{2} R$ WHY: the extra $R$ is the expansion-work tax when pressure is held fixed. $γ = \frac{C _{p}}{C _{v}} = \frac{\frac{5}{2} R}{\frac{3}{2} R} = \frac{5}{3} \approx 1.67$ In numbers: $C_{p} = \frac{5}{2} (8.314) = 20.79 J mol^{- 1} K^{- 1}$ .

Exercise 2

A gas has $C_{p} = 29.10 J mol^{- 1} K^{- 1}$ . Find $C_{v}$ .

Recall Solution (L1.2)

Mayer's relation rearranged to isolate $C_{v}$ : $C_{v} = C_{p} - R = 29.10 - 8.314 = 20.79 J mol^{- 1} K^{- 1}$ WHY subtract $R$ : the constant-volume case does no work, so it needs $R$ less heat per degree.

L2 — Application

Exercise 3

$3$ moles of an ideal monatomic gas are heated from $300 K$ to $400 K$ at constant volume. Find the heat supplied.

Recall Solution (L2.1)

WHAT: constant volume $\Rightarrow$ no work $\Rightarrow$ all heat becomes $Δ U$ . $Q_{v} = n C_{v} Δ T = 3 \times \frac{3}{2} (8.314) \times (400 - 300)$ $Q_{v} = 3 \times 12.471 \times 100 = 3741.3 J$ WHY $C_{v}$ and not $C_{p}$ : with $d V = 0$ , the first law gives $δ Q = d U$ , and $Δ U = n C_{v} Δ T$ .

Exercise 4

Same gas, same $n = 3$ , same $300 \to 400 K$ , but now at constant pressure. Find $Q_{p}$ and the expansion work $W$ .

Recall Solution (L2.2)

First get $C_{p}$ : $C_{p} = \frac{3}{2} R + R = \frac{5}{2} R$ . $Q_{p} = n C_{p} Δ T = 3 \times \frac{5}{2} (8.314) \times 100 = 3 \times 20.785 \times 100 = 6235.5 J$ The work is the difference between the two heats (same $Δ U$ in both): $W = Q_{p} - Q_{v} = 6235.5 - 3741.3 = 2494.2 J$ Cross-check with $W = n R Δ T = 3 \times 8.314 \times 100 = 2494.2 J$ ✓. Look at the bar chart — the mint slice is exactly this work.

Figure — Heat capacities Cp, Cv; relationship Cp − Cv = nR (ideal gas)

L3 — Analysis

Exercise 5

An ideal gas is heated and found to need $2910 J$ at constant pressure but only $2077 J$ at constant volume for the same $Δ T$ and same $n$ . Identify $γ$ and hence the atomicity.

Recall Solution (L3.1)

The ratio of heats at equal $n Δ T$ is the ratio of heat capacities: $γ = \frac{Q _{p}}{Q _{v}} = \frac{C _{p}}{C _{v}} = \frac{2910}{2077} = 1.401$ WHY this works: $Q_{p} = n C_{p} Δ T$ , $Q_{v} = n C_{v} Δ T$ ; the common $n Δ T$ cancels. $γ \approx 1.40 = 7/5$ ⇒ diatomic (rotational modes active, vibration frozen). See Degrees of Freedom.

Exercise 6

For that same experiment ( $n$ moles, $Δ T$ ), $Q_{p} - Q_{v} = 833 J$ . If $Δ T = 50 K$ , find $n$ .

Recall Solution (L3.2)

The gap between the two heats is pure expansion work: $Q_{p} - Q_{v} = n R Δ T$ Solve for $n$ : $n = \frac{Q _{p} - Q _{v}}{R Δ T} = \frac{833}{8.314 \times 50} = \frac{833}{415.7} = 2.004 \approx 2 mol$ WHY: $Q_{p} - Q_{v} = n C_{p} Δ T - n C_{v} Δ T = n (C_{p} - C_{v}) Δ T = n R Δ T$ .

L4 — Synthesis

Exercise 7

$1$ mole of a diatomic ideal gas ( $C_{v} = \frac{5}{2} R$ ) is taken from state A $(300 K)$ to state B $(500 K)$ . Compute $Δ U$ and $Δ H$ — and argue why these two answers do not depend on whether the path was isobaric, isochoric, or something curvy in between.

Recall Solution (L4.1)

$Δ U = n C_{v} Δ T = 1 \times \frac{5}{2} (8.314) \times 200 = 4157 J$ $Δ H = n C_{p} Δ T = 1 \times \frac{7}{2} (8.314) \times 200 = 5819.8 J$ ( $C_{p} = \frac{5}{2} R + R = \frac{7}{2} R$ .) WHY path-independent: $U$ and $H$ are state functions — for an ideal gas both depend only on $T$ . Any path from $300$ to $500 K$ gives the same endpoints, hence the same $Δ U$ , $Δ H$ . (See Internal Energy, Enthalpy.) Only $Q$ and $W$ care about the path.

Exercise 8

The gas of Exercise 7 is instead expanded adiabatically ( $Q = 0$ ) so its temperature falls from $500 K$ back to $300 K$ . How much work does the gas do?

Recall Solution (L4.2)

WHAT: adiabatic ⇒ $Q = 0$ , so the first law $Δ U = Q - W$ gives $W = - Δ U$ . Here $Δ T = 300 - 500 = - 200 K$ : $Δ U = n C_{v} Δ T = 1 \times \frac{5}{2} (8.314) \times (- 200) = - 4157 J$ $W_{by gas} = - Δ U = + 4157 J$ WHY positive: the gas cools by spending its own internal energy as work — nothing came in as heat. Contrast Isothermal vs Adiabatic Processes: an isothermal expansion would keep $T$ fixed and take that energy from a reservoir instead.

L5 — Mastery

Exercise 9

A mixture contains $2 mol$ of a monatomic gas ( $C_{v, 1} = \frac{3}{2} R$ ) and $3 mol$ of a diatomic gas ( $C_{v, 2} = \frac{5}{2} R$ ). Find the effective $C_{v}$ , $C_{p}$ and $γ$ of the mixture (per mole of mixture).

Recall Solution (L5.1)

WHAT: heat capacities add — total internal-energy change is the sum of each component's. $C_{v, mix} = \frac{n _{1} C _{v, 1} + n _{2} C _{v, 2}}{n _{1} + n _{2}} = \frac{2 ( \frac{3}{2} R ) + 3 ( \frac{5}{2} R )}{5} = \frac{3 R + 7.5 R}{5} = \frac{10.5 R}{5} = 2.1 R$ Numerically $C_{v, mix} = 2.1 \times 8.314 = 17.46 J mol^{- 1} K^{- 1}$ . Mayer still holds for the (ideal) mixture: $C_{p, mix} = C_{v, mix} + R = 3.1 R = 25.77 J mol^{- 1} K^{- 1}$ $γ_{mix} = \frac{C _{p, mix}}{C _{v, mix}} = \frac{3.1 R}{2.1 R} = \frac{3.1}{2.1} = 1.476$ WHY Mayer survives mixing: each component is ideal, so the whole mixture obeys $P V = n R T$ , and the $n R Δ T$ expansion tax is the same idea for the pooled gas.

Exercise 10

A rigid vessel and a piston vessel each hold $1 mol$ of the same ideal gas at $300 K$ . Both receive exactly $1000 J$ of heat. The gas has $C_{v} = \frac{5}{2} R$ . Find the final temperature in each vessel, and explain the temperature gap using the picture.

Recall Solution (L5.2)

Rigid (constant V): all heat → internal energy. $Δ T_{v} = \frac{Q}{n C _{v}} = \frac{1000}{1 \times \frac{5}{2} ( 8.314 )} = \frac{1000}{20.785} = 48.11 K \Rightarrow T = 348.11 K$ Piston (constant P): part of the heat leaks as work, so temperature rises less. $Δ T_{p} = \frac{Q}{n C _{p}} = \frac{1000}{1 \times \frac{7}{2} ( 8.314 )} = \frac{1000}{29.099} = 34.37 K \Rightarrow T = 334.37 K$ WHY the gap: for the same $Q$ , the piston vessel spends some energy pushing outward, leaving less to heat the molecules. Look at figure s02 — the coral bar (constant $P$ ) rises lower because the mint segment escaped as work. Difference $\approx 13.7 K$ .

Recall One-line self-check

Why is $γ$ always greater than 1? ::: Because $C_{p} = C_{v} + R$ with $R > 0$ , so $C_{p} > C_{v}$ , hence $C_{p} / C_{v} > 1$ . Does $Δ U = n C_{v} Δ T$ hold in an isobaric process? ::: Yes — for an ideal gas it holds in every process.

Connections

First Law of Thermodynamics — the $Δ U = Q - W$ engine behind every solution.
Ideal Gas Law — gives $P Δ V = n R Δ T$ , the expansion-work term.
Adiabatic Process, Isothermal vs Adiabatic Processes — Exercise 8.
Degrees of Freedom — sets $C_{v}$ and hence $γ$ (Exercise 5).
Real Gases (van der Waals) — where Mayer's simple $R$ gets corrected.