Piston displacement

Classical Mechanics Level 5

The device consists of a piston and cylinder with air of mass m m contained within it. The cylinder and piston are perfectly insulated, therefore there is no heat transfer between system and surroundings.

A Block weighing W = 1000 N W=1000 \text{ N} is placed on the piston and the system is allowed to come to equilibrium. Now Q = 1 KJ Q=1 \text{ KJ} amount of heat is added to the system due to which the piston moves up by a distance h h but this happens very slowly (Quasi-equilibrium process).

Case 1: The device is kept in vacuum. P atm = 0 KPa { P }_{ \text{atm} }=0 \text{ KPa}
Let the distance moved by the piston in this case be h 1 { h }_{ 1 } .

Case 2: The device is kept in a place where P atm = 100 KPa { P }_{\text{atm} }=100 \text{ KPa} Let the distance moved by the piston in this case be h 2 { h }_{ 2 } .

Evaluate h 1 − h 2 { h }_{ 1 }-{ h }_{ 2 } in cm \text{cm} .

Details and Assumptions :

  1. Area of the piston = 150 cm 2 150 \text{ cm }^{ 2 } .
  2. Heat capacity ratio of air γ \gamma = c p c v = 1.4 \frac { { c }_{ p } }{ { c }_{ v } } =1.4 , c p { c }_{ p } and c v { c }_{ v } indicates specific heat at constant pressure and volume respectively.
  3. Piston weight is negligible.
  4. Air can be treated as an ideal gas.


The answer is 17.14.

Rahul Badenkal by Rahul Badenkal , 26, India

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2 solutions

Rahul Badenkal
Jul 22, 2015

It is given that the piston moves very slowly. So at any instant the the net force acting on the piston is zero. Therefore, the pressure of the gas will always be equal to the outside pressure which is a constant in both the cases. Hence, It is a constant pressure process

Case 1: P = W A P=\frac { W }{ A }

Q = m c p ( T 2 − T 1 ) { Q }=m{ c }_{ p }({ T }_{ 2 }-{ T }_{ 1 })

From ideal gas equation ( P V = m R g a s T ) (PV=m{ R }_{ gas }T)\quad :

A t s t a t e 1 T 1 = W A V 1 m R a i r = W V 1 A m R a i r At\quad state\quad 1\quad \quad \quad \quad { T }_{ 1 }=\frac { \frac { W }{ A } { V }_{ 1 } }{ m{ R }_{ air } } =\frac { W{ V }_{ 1 } }{ Am{ R }_{ air } }

A t s t a t e 2 T 2 = W A V 2 m R a i r = W V 2 A m R a i r At\quad state\quad 2\quad \quad \quad \quad { T }_{ 2 }=\frac { \frac { W }{ A } { V }_{ 2 } }{ m{ R }_{ air } } =\frac { W{ V }_{ 2 } }{ Am{ R }_{ air } }

Q = m c p W A m R a i r ( V 2 − V 1 ) , V 2 − V 1 = Δ V = A h 1 Q=m{ c }_{ p }\frac { W }{ Am{ R }_{ air } } ({ V }_{ 2 }-{ V }_{ 1 }),\quad \quad { V }_{ 2 }-{ V }_{ 1 }=\Delta V=A{ h }_{ 1\\ \\ }

Q = m c p W A m R a i r ( A h 1 ) Q=m{ c }_{ p }\frac { W }{ Am{ R }_{ air } } (A{ h }_{ 1\\ \\ })

h 1 = 1 W . R a i r c p Q , c p = γ γ − 1 R a i r { h }_{ 1\\ \\ }=\frac { 1 }{ W } .\frac { { R }_{ air } }{ { c }_{ p } } Q,\quad \quad \quad \quad { c }_{ p }=\frac { \gamma }{ \gamma -1 } { R }_{ air }

h 1 = 1 W . γ − 1 γ Q , { h }_{ 1\\ \\ }=\frac { 1 }{ W } .\frac { \gamma -1 }{ \gamma } Q,\quad

h 1 = 0.2857 m o r 28.57 c m { h }_{ 1\\ \\ }=0.2857m\quad or\quad 28.57cm

Replacing W a s W + P a t m . A W\quad as\quad W+{ P }_{ atm }.A for case 2, as the net force is weight + Atmospheric force.

h 2 = 1 W + P a t m . A . γ − 1 γ Q , { h }_{ 2 }=\frac { 1 }{ W+{ P }_{ atm }.A } .\frac { \gamma -1 }{ \gamma } Q,

h 2 = 0.1143 m o r 11.43 c m { h }_{ 2 }=0.1143m\quad or\quad 11.43cm

h 1 − h 2 = 17.14 c m { h }_{ 1 }-{ h }_{ 2\\ \\ }=17.14cm

4 Helpful 0 Interesting 0 Brilliant 0 Confused

My solution is basically the same as yours, except everywhere you used the mass "m" I think you should have used "n" (the number of moles).

The ideal gas law is P V = n R T PV=nRT and the definition of the (molar) heat capacity is C ≡ Q n Δ T C\equiv\frac{Q}{n\Delta T} (where Q Q is heat needed to raise temperature by Δ T \Delta T ).

(It doesn't affect the answer though, because it cancels out either way.)

Nathanael Case - 5 years, 10 months ago

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P V = n R u T PV=n{ R }_{ u }T , Where:

n = n= no. of moles = m M =\frac { m }{ M } , m = m= given mass, M = M= Molecular mass

R u = { R }_{ u }= Universal gas constant = 8.314 J / k g . K =8.314J/kg.K

P V = m M R u T PV=\frac { m }{ M } { R }_{ u }T

P V = m R u M T PV=m\frac { { R }_{ u } }{ M } T

P V = m R g a s T , PV=m{ R }_{ gas }T, , where R g a s { R }_{ gas } value depends on the type of gas being used.

That's why I wrote R a i r { R }_{ air } everywhere in the solution instead of just writing R R

Rahul Badenkal - 5 years, 10 months ago

Okey because process is slow and the piston is 'massless' the net force on him must be zero:

P = c o n s t P=const so Δ P = 0 \Delta P=0 the process is isobaric.

P V = n R T PV=nRT differentiating this you will get

P Δ V = n R Δ T P \Delta V=nR \Delta T

Q = C p n Δ T = C p P Δ V Q=C_p n \Delta T=C_p P \Delta V

P = F n e t S P=\frac{F_{net}}{S}

And

Δ V = S Δ h \Delta V=S \Delta h

C p C p − 1 = γ = 1.4 \frac{Cp}{Cp-1}=\gamma=1.4

now

C p = 3.5 R Cp=3.5R

C a s e 1 : F n e t = W , Δ h = h 1 = 28.57 c m Case 1: F_{net}=W, \Delta h=h_1=28.57cm

C a s e 2 : F n e t = W + P a t m S , Δ h = h 1 = 11.43 c m Case 2: F_{net}=W+P_{atm}S, \Delta h=h_1=11.43cm

P.S. I think it's overrated...

1 Helpful 0 Interesting 0 Brilliant 0 Confused

I agree. It should be a level 3 or at max a level 4 problem.

Rahul Badenkal - 5 years, 8 months ago

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