[Methyl_Manic]

Hi Guys,
i know this may be an elemenary question for some, and i understand the question, i just dont understand how to go about calculation:

the question is:
for an ideal gas, during an isothermal expansion for A->C, Calculate w and q for the isothermal pathway (AC),
(i was given a p-V diagram where at state A: p =100000Pa and V = 0.05m³, and state C: p =10000Pa and V = 0.1m³  and an isotherm  linking them. i was given equations: pV = nRT and U =(3/2)RT

i understand that the work can be calculated for this process as:

w = RTln(Vf/Vi)

but what i dont understand is how i calculate T? from what i understand i would use the ideal gas law pV = nRT and it is assumed that n=1. so can i use the ideal gas law for this using p and V at either state A or state C to work out T, as the process is isothermal so dT = 0, dU = 0 and also q != 0. we can use the first law to calculate q as the internal energy U does not change? and i was given the equation U =(3/2)RT for this gas:

i calculate T using pV/nR = T = (100000x0.05)/8.314 =  601.39 K,

putting this into w = RTln(Vf/Vi) i get work as:

w = 8.314*601.39*ln(0.1/0.05) = 3465.7 J

and thus U = (3/2)RT = (3/2)*8.314*601.39 = 7500 J

using this in the first law: U = q + w

q = U - w = 7500 - 3465.7 = 4034.3 J

so heat had to be added to the system to maintain the internal energy of the system in this isothermal process. and positive work was done on the surrounds in the process.

is this correct?

Thanks!

[mjc123]

i was given a p-V diagram where at state A: p =100000Pa and V = 0.05m3, and state C: p =10000Pa and V = 0.1m3  and an isotherm  linking them.

This is not isothermal. Along an isotherm PV is constant (= nRT). You are doubling the volume while the pressure drops by a factor of 10. Are you sure you have copied the data correctly?

[Methyl_Manic]

Hi,
thankyou for the reply. i have attached a link to the question. this is from a final examination. the question states that the system move along an isothermic pathway from state A to state C, an expansion of the gas at constant temperature.

[img=http://s15.postimg.org/ol6gg2407/Screen_Shot_2014_10_24_at_11_28_43_pm.jpg]

[mjc123]

Ah. At C, P = 50000 Pa (not 10000).
Actually w = -nRTln(V_f/V_i). nRT = PV = constant. You don't need to know n and T separately. w is the work done on the gas (not by the gas), hence the negative sign. Your calculation looks correct but needs a -.
q = ΔU - w. As you correctly say, U does not change in an isothermal process, so q = -w. q is positive as heat is put into the gas.

[Methyl_Manic]

Ah you are quite correct. this is me not copying the question correctly.

for me there seems to be a little bit of confusion about the sign of work. upon a little research i have found that this confusion may be as a result of the first law being written in terms of physics and chemistry text books. where The physics text writes U=q−w while chemistry says U=q+w.

from a physics perspective work on by the system is positive and work done on the system is negative. and vice versa for chemistry. are both forms still acceptable if explained? i.e. i will loose marks if i donot write work as being negative.. but what is this negative work with respect to? the system? the system done negative work? (makes no sense) or the surroundings?  this is where i start to get to two contradictory explanations for work and thus confusion. the only way i can think of it (in the chemistry form) as the available work of the system has now decreased as it did some work on the surroundings and this is why it is negative with respect to the system.

sorry this may be simple.. but it seems like one could look at this two ways.

[mjc123]

Ah, physicists - cause of all the problems in the world (apart from those caused by lawyers, bankers, Simon Cowell etc...)
I can't speak for physicists, but chemists take the point of view of the system (not surroundings), and anything that increases the energy content of the system is positive (e.g. heat input to the system, q, or work done on the system, such as by compression, w). For an isothermal process ΔU = 0, so q = -w; you have to put heat in (positive q) to make the gas expand (negative w); or compressing the gas (positive w) generates heat which is lost to the surroundings (negative q). For an adiabatic process q = 0 and ΔU = w, so compressing a gas causes it to heat up and expanding it cools it down.

the only way i can think of it (in the chemistry form) as the available work of the system has now decreased as it did some work on the surroundings and this is why it is negative with respect to the system.

That sounds a good explanation. I also agree with your suggestion to be explicit about what work you're referring to, e.g. saying "work done on system = -X J" rather than simply "work = -X J". Then you oughtn't to be penalised even if your teacher was looking for work done by system = +X J.