Thermodynamics - Question with my answer

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    Thermodynamics
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Homework Help Overview

The discussion revolves around a thermodynamics problem involving the compression of a gas, where the gas transitions from an initial volume to a final volume under a specified external pressure, while releasing heat. Participants are analyzing the implications of work done on the system and the changes in internal energy.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculations related to work done during gas compression and the corresponding changes in internal energy. Questions arise regarding the interpretation of energy transfer and the signs associated with work and heat.

Discussion Status

There is an active exploration of the relationships between work, heat, and internal energy. Some participants provide calculations and reasoning, while others seek clarification on the implications of energy flow into or out of the system. Multiple interpretations of the signs of work and heat are being examined.

Contextual Notes

Participants are navigating the definitions and signs of work and heat in the context of thermodynamic processes, with specific attention to the implications of positive and negative values in their calculations.

manal950
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A gas is compressed from an initial volume of 5.40 L to a final volume of 1.23 L by an external pressure of 1.00 atm . During the compression the gas releases 128 J of heat.

My answer :

given data :
v1 = 5.40 l
v2 = 1.23 l
P1 - 1.00

now convert L to m^3 and atm to pa so
1 l =0.001 cubic meters
1 atm = 101,325 pa
v1 = 5.4X10^-3 m^3
v2 = 1.23 X 10^-3 m^3
p = 101325 pa

now DU = -Q - W
w = -pXDv = 101325(1.23 X 10^-3 - 5.4X10^-3 )
w = 422.5 J

now DU = -Q + W
= -128 + 422.5
Du = 294.5 J
 
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W = p ΔV
= - 423 J
The negative work being done means that it is receiving energy from an external source (doing work on it)
Q = -128 J
The negative means that it is losing energy (heat)
so that from
ΔU = Q - W
= -128 -(-423)
= 295 J
the internal energy increased by this amount
 
thanks, but how we can no from the question that the system receiving energy from an external source ...?
 
Work and energy go hand in hand. The system will loose energy if it is doing positive work and gain energy if it is doing negative work. It is doing positive work if it is expanding, since it is exerting an outwards force against the atmosphere while expanding, negative work if contracting, since the system is still exerting an outwards force on the atmosphere, but now it is motioning in the opposite direction to the force, that is contracting inwards. Also if ΔU is positive then we have a net energy flow into the system. We also need to consider the heat flow, which is a form of energy.
 
Last edited:

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