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Monatomic ideal gas expansion

  • Thread starter scholio
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  • #1
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Homework Statement



one mole of a monatomic gas expands from V_i = V_0 to V_f = 3V_0 according to the relationship:

p = P_0[1 - 0.9sin( (pi(V - V_0))/4V_0 )]

a) compute the work done by the gas as it expands from V_i = V_0 to V_f = 3V_0

b) compute the net heat flow during this expansion

c) does heat flow into or out of the gas during this expansion?


Homework Equations



PV = nRT where P is pressure, V is volume, n is moles, R = 8.314 constant, T is temperature

Q = W = nRT ln (V_2/V_1) where W is work, Q is heat, V_2 and V_1 is volume

not sure what else

The Attempt at a Solution



I haven't got far because i don't understand the given relationship, what is the difference between the capital and lowercase p, is the lowercase p stand for momentum? i doubt it..

i used the second eq and tried to solve for work but didn't know how to implement the relationship given, where does it come into play?

any help appreciated...
 

Answers and Replies

  • #2
Andrew Mason
Science Advisor
Homework Helper
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325

Homework Statement



one mole of a monatomic gas expands from V_i = V_0 to V_f = 3V_0 according to the relationship:

p = P_0[1 - 0.9sin( (pi(V - V_0))/4V_0 )]

a) compute the work done by the gas as it expands from V_i = V_0 to V_f = 3V_0

b) compute the net heat flow during this expansion

c) does heat flow into or out of the gas during this expansion?


Homework Equations



PV = nRT where P is pressure, V is volume, n is moles, R = 8.314 constant, T is temperature

Q = W = nRT ln (V_2/V_1) where W is work, Q is heat, V_2 and V_1 is volume

not sure what else

The Attempt at a Solution



I haven't got far because i don't understand the given relationship, what is the difference between the capital and lowercase p, is the lowercase p stand for momentum? i doubt it..
Ignore lowercase. p is just the pressure P.

Start with the definition of work:

[tex]W = \int_{V_0}^{3V_0} Pdv[/tex]

Substitute for P:

[tex]W = \int_{V_0}^{3V_0} P_0(1 - \sin(\pi(V - V_0)/4V_0 )dv[/tex]

AM
 
  • #3
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thanks AM, so does it become:

W = P_0[1 - 0.9sin( (pi(V - V_0))/4V_0 )](3V_0) - P_0[1 - 0.9sin( (pi(V - V_0))/4V_0 )](V_0) ?

how does the V and the V_0 come into play within the given function?

cheers
 
  • #4
160
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thanks AM, so does it become:

W = P_0[1 - 0.9sin( (pi(V - V_0))/4V_0 )](3V_0) - P_0[1 - 0.9sin( (pi(V - V_0))/4V_0 )](V_0) ?

how does the V and the V_0 come into play within the given function?

cheers
sorry double posting, 'edit' option wasn't available for some reason.
part b asks to determine the heat flow during the expansion, am i correct to assume that the pressure changes as does the volume, thus the expansion is isothermic

and should use eq Q_isotherm = W = nRT ln (V_2/V_1) where n = 1 mole, R = constant 8.314, T is not specified in problem, assumed constant and V_2 = 3V_0, V_1 = V_0

plugging in those values, for Q (heat) i got Q = 9.13T where T is the temp, constant

part c, asks if heat flows into or out of the gas during the expansion --> i said, since Q = 9.13T is positive, it means heat is flowing into the gas

correct??
 
Last edited:
  • #5
Andrew Mason
Science Advisor
Homework Helper
7,563
325
thanks AM, so does it become:

W = P_0[1 - 0.9sin( (pi(V - V_0))/4V_0 )](3V_0) - P_0[1 - 0.9sin( (pi(V - V_0))/4V_0 )](V_0) ?

how does the V and the V_0 come into play within the given function?

cheers
You just have to figure out the integral:

[tex]W = \int_{V_0}^{3V_0} P_0(1 - \sin(\pi(V - V_0)/4V_0 )dv[/tex]

which is of the form:

[tex]W = \int_{V_0}^{3V_0} P_0dv - \int_{V_0}^{3V_0} P_0\sin(kV - \phi) dv[/tex]

Work that out. Remember [itex]P_0[/itex] and [itex]V_0[/itex] are constants.

AM
 

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