Calculating Work for Monatomic Gas Expansion

[scholio]

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...

 

[Andrew Mason]

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:

W = \int_{V_0}^{3V_0} Pdv

Substitute for P:

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

AM

 

[scholio]

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

 

[scholio]

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??

 

[Andrew Mason]

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:

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

which is of the form:

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

Work that out. Remember P_0 and V_0 are constants.

AM