Physics Thermodynamics - Finding Work

In summary, the problem involves a 1.20 mol sample of an ideal diatomic gas with initial conditions of 1.20 atm and 380 K, undergoing a process where pressure increases linearly with temperature. The final conditions are 680 K and 1.83 atm. The question asks for the work done by the gas, but there seems to be inconsistencies in the given equations. Two possible interpretations of "linear" result in different equations for work.
  • #1
misa
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Homework Statement


A 1.20 mol sample of an ideal diatomic gas at a pressure of 1.20 atm (P1) and temperature of 380 K (T1) undergoes a process in which its pressure increases linearly with temperature. The final temperature and pressure are 680 K (T2) and 1.83 atm (P2).

(b) Determine the work done by the gas.


Homework Equations



(T2 - T1) / (P2 - P1) = constant

V2 / V1 = P1T2 / P2T1

PV = nRT

W = integral of PV

The Attempt at a Solution



The answer is supposed to be [nR(P1T2 - P2T1) ln(P2/P1)] / (P2 - P1)
but I can't figure out how to get there despite the hours I have poured into this problem. All I know is that volume is not constant in order for there to be work done. Also, I have the vague idea of finding the linear equation or relationship between T and P using that equation to plug in T in PV = nRT. Then, technically, there would be a graph for PV and I could solve for V1 and V2, which can be the limits of integration for the integral of PV (to get work). But isn't that way too complicated, especially as an integral...so I was wondering whether someone else has any idea.

Please help explain, and thank you!
 
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  • #2
I certainly understand your frustration!
If P varies linearly with T, you would expect P = kT. Put that in V = nRT/P and you get that V is a constant, which is wrong. Looks like the question is inconsistent!

The other meaning of "linear" would be P = kT + b so V = nRT/(kT+b) which doesn't seem likely. It does give quite a complex answer.
Since P and V both vary, dW = PdV + VdP
I worked the integral quickly and got
W = -nRb(P2 - P1) + nR/k*[P2-P1 -ln(P2/P1)]
Could well be errors in that! It doesn't look much like the given answer, does it?
 
  • #3
Ah, thank you so much for the insight. (I'll have to try and work that out when I can find the time.)
 

Related to Physics Thermodynamics - Finding Work

What is thermodynamics?

Thermodynamics is the branch of physics that deals with the study of heat and its relationship to energy and work. It also encompasses the study of how energy is transferred and transformed between different forms, such as heat, work, and internal energy.

What is work in thermodynamics?

In thermodynamics, work is defined as the process of exerting a force over a distance to move an object. It is a form of energy transfer and is typically measured in joules (J). In the context of thermodynamics, work is often associated with the expansion or compression of a gas.

How is work calculated in thermodynamics?

The work done in thermodynamics is calculated using the formula W = F x d, where W is work, F is the force applied, and d is the distance over which the force is applied. This formula is used to calculate the work done in various thermodynamic processes, such as isobaric, isothermal, and adiabatic processes.

What is 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. This means that the total energy of a system remains constant. In other words, the change in the internal energy of a system is equal to the heat added to the system minus the work done by the system.

What are some real-life examples of thermodynamics?

Some real-life examples of thermodynamics include the operation of a car engine, the heating and cooling of a home, and the functioning of a refrigerator. In these examples, thermodynamics principles are used to transfer and transform energy, such as converting heat into work or vice versa.

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