Thermodynamics work done question

In summary, the volume of one mole of air is V = nRT/P. The mass of one mole of air is M = 29 g/mole. The work done in expanding air from 1M Pa to 200kPa is W = nRT/P. Therefore, W = nRT/P*M = nRT/29 = 6.7 kJ/kg.
  • #1
junglep
8
0
hey guys got this question that i have been stuck on for a while.

air is expanded from 1M Pa at 327 degrees celcius to 200kPa in a closed piston cylinder device. for the process PV^1.2 = constant. calculate the work done in kJ/kg during this process

i hav managed to work out the temperature after expansion using T2/T1 = (P1/P2)^(n-1/n) but i don't know how to work out the work done without knowing the mass or any of the volumes

if work = (p1v1 - p2v2)/ 1-n

then surely i need the volumes to work out the work done

any help will be welcomed

cheers
 
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  • #2
If the gaz can be considered ideal, then by conservation of the number of moles of gaz,

[tex]\nu_i = \nu_f[/itex]

you must have

[tex]V_f=\frac{p_fT_i}{p_iT_f}V_i[/tex]

So

[tex]W=\int_{V_i}^{\frac{p_fT_i}{p_iT_f}V_i}pdV = \int_{V_i}^{\frac{p_fT_i}{p_iT_f}V_i} \frac{\alpha}{V^{1.2}}dV[/tex]

And substitude back [itex]\alpha = p_iV_i^{1.2}[/itex] at the end.
 
  • #3
but i don't know any of the volumes so this method would not work
 
  • #4
junglep said:
hey guys got this question that i have been stuck on for a while.

air is expanded from 1M Pa at 327 degrees celcius to 200kPa in a closed piston cylinder device. for the process PV^1.2 = constant. calculate the work done in kJ/kg during this process

i hav managed to work out the temperature after expansion using T2/T1 = (P1/P2)^(n-1/n) but i don't know how to work out the work done without knowing the mass or any of the volumes

if work = (p1v1 - p2v2)/ 1-n

then surely i need the volumes to work out the work done

any help will be welcomed

cheers
If [itex]PV^\alpha = K[/itex] where [itex]\alpha = 1.2[/itex] (note: this is not the [itex]\gamma[/itex] for air which is 1.4), then substituting V = nRT/P gives:

[tex]P^{1-\alpha}T^\alpha = K/n^\alpha R^\alpha = K'[/tex]

So:

[tex]P_1^{1-\alpha}T_1^\alpha = P_2^{1-\alpha}T_2^\alpha = K'[/tex]

From that, work out PdV in terms of K' and T and integrate from T1 to T2

AM
 
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  • #5
If you succed, would you please post the answer junglep?
 
  • #6
quasar987 said:
If you succed, would you please post the answer junglep?
Find T2 from the relationship:

[tex]P_1^{(1-\alpha)}T_1^\alpha = P_2^{(1-\alpha)}T_2^\alpha[/tex]

so:

[tex]T_2 = \left(P_1^{(1-\alpha)}T_1^\alpha/P_2^{(1-\alpha)}\right)^{1/\alpha}[/tex]

Use PV=nRT to find V:

[tex]V_1 = nRT_1/P_1[/tex]

[tex]V_2 = nRT_2/P_2[/tex]

Integrating PdV from V1 to V2 using [itex]P = K/V^\alpha[/itex]:

[tex]W = \int_{V_1}^{V_2} PdV = \int_{V_1}^{V_2} KdV/V^\alpha[/tex]

You just have to work that out.

AM
 
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  • #7
what is n in the equation

pv = nRT?

i thought the perfect gas eqn was pv = mass * R * T

also i am not given a value for the gas constant (R). if it is any help the answer that is given in the book is in kJ/kg not in J.
 
  • #8
junglep said:
what is n in the equation

pv = nRT?

i thought the perfect gas eqn was pv = mass * R * T

also i am not given a value for the gas constant (R). if it is any help the answer that is given in the book is in kJ/kg not in J.
n is the number of moles of the gas. R is in units of J/mole K.

This problem does not give you n or V, so assume n = 1 in which case: PV = MRT where M is the mass of one mole of air (29 g/mole). Essentially, you are working out and using the volume for one mole of air.

AM
 

1. What is the definition of work in thermodynamics?

In thermodynamics, work is defined as the transfer of energy from one system to another due to a force acting on the system. This can be expressed mathematically as W = F * d, where W is work, F is the force applied, and d is the displacement of the system.

2. How is work related to heat in thermodynamics?

Work and heat are both forms of energy and can be interchanged in thermodynamic processes. However, work is a result of energy transfer due to a force, while heat is a result of energy transfer due to a temperature difference.

3. What are the different types of work in thermodynamics?

In thermodynamics, there are three main types of work: mechanical work, electrical work, and chemical work. Mechanical work involves the application of a force to cause a displacement, electrical work involves the movement of charged particles, and chemical work involves the formation or breaking of chemical bonds.

4. How is work calculated in thermodynamics?

The calculation of work in thermodynamics depends on the type of work being done. For mechanical work, it can be calculated using the equation W = F * d. For electrical work, it can be calculated using the equation W = V * I * t, where V is voltage, I is current, and t is time. For chemical work, it can be calculated using the equation W = -∆G, where ∆G is the change in Gibbs free energy.

5. How is work done related to the first law of thermodynamics?

The first law of thermodynamics states that energy cannot be created or destroyed, only transferred or converted from one form to another. Work done is a form of energy transfer, so it is closely related to the first law of thermodynamics. In fact, the first law can be expressed as ∆U = Q - W, where ∆U is the change in internal energy, Q is heat added to the system, and W is work done by the system.

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