Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 1 / 36C (Statistical Physics)

Table of contents
- a 36C
  - Solution a
- b 36C
  - Solution b
- c 36C
  - Solution c
- d 36C
  - Solution d

a

Solution

For a fixed number

N

of particles, the ideal gas equation of state is

p V = N k_{B} T .

Its internal energy depends only on temperature, and

d E = C_{V} d T .

When the heat capacity at constant volume is constant, a choice of the energy zero gives

E = C_{V} T .

b

Solution

At constant volume, the First law of thermodynamics gives

δ Q = d E = C_{V} d T .

At constant pressure,

δ Q = C_{p} d T = d E + p d V .

Differentiating

p V = N k_{B} T

at fixed

p

gives

p d V = N k_{B} d T .

Consequently,

C_{p} d T = (C_{V} + N k_{B}) d T,

and the Mayer relation is

C_{p} - C_{V} = N k_{B} .

The specific-heat ratio is therefore

γ = \frac{C _{p}}{C _{V}} = 1 + \frac{N k _{B}}{C _{V}} .

c

Solution

An adiabatic process exchanges no heat with its surroundings, so

δ Q = 0

. For a reversible process it is also isentropic. The first law then gives

C_{V} d T = d E = - p d V .

Using

p = N k_{B} T / V

and the Mayer relation,

\frac{d T}{T} = - \frac{N k _{B}}{C _{V}} \frac{d V}{V} = - (γ - 1) \frac{d V}{V} .

Integration yields

T V^{γ - 1} = constant .

Substituting

T = p V / (N k_{B})

gives the reversible ideal-gas adiabat

p V^{γ} = constant .

d

Solution

In the

(p, V)

plane, the Otto cycle consists of:

$A \to B$ : an adiabatic compression along $p V^{γ} = constant$ from $V_{1}$ to $V_{2}$ ;
$B \to C$ : a vertical constant-volume rise at $V_{2}$ ;
$C \to D$ : an adiabatic expansion back to $V_{1}$ ;
$D \to A$ : a vertical constant-volume fall at $V_{1}$ .

It is a clockwise loop, so its enclosed area is the positive net work output. In the

(T, S)

plane, the two reversible adiabats

A \to B

and

C \to D

are vertical constant-entropy segments. Constant-volume heating

B \to C

moves upward and to the right, while constant-volume cooling

D \to A

moves downward and to the left.

Let

r = \frac{V _{1}}{V _{2}} > 1.

The two adiabatic relations give

T_{B} = T_{A} r^{γ - 1}, T_{C} = T_{D} r^{γ - 1} .

Because heat is exchanged at constant volume,

Q_{1} = C_{V} (T_{C} - T_{B}), Q_{2} = C_{V} (T_{D} - T_{A}),

where

Q_{2}

denotes the positive amount rejected. Hence

\frac{Q _{2}}{Q _{1}} = \frac{T _{C} / r ^{γ - 1} - T _{A}}{T _{C} - T _{A} r ^{γ - 1}} = \frac{1}{r ^{γ - 1}} .

The thermal efficiency is therefore

η = \frac{W}{Q _{1}} = 1 - \frac{Q _{2}}{Q _{1}} = 1 - \frac{1}{r ^{γ - 1}} .

For a fixed gas,

γ

is fixed and the idealized efficiency increases monotonically with the compression ratio

r

, so it is maximized by making

r

as large as the physical constraints allow.

 Ancestors (9)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version