One kg of air is heated in a closed rigid vessel

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Homework Help Overview

The problem involves heating one kilogram of air in a closed rigid vessel, with a temperature change from 27°C to 427°C. Participants are tasked with finding the heat transferred and the change in internal energy, using given values for specific gas constants.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the heat transfer equation but questions the unit consistency for R and Cv. Other participants discuss the implications of constant volume on work done and relate it to the change in internal energy.

Discussion Status

Participants are exploring different interpretations of the problem, with some providing guidance on the relationship between heat transfer and internal energy in a rigid container. There is a lack of consensus on the approach to calculating work and heat transfer.

Contextual Notes

Assumptions regarding the properties of air and the conditions of the closed system are under discussion, particularly the implications of constant volume on work and energy calculations.

manal950
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Homework Statement



One kg of air is heated in a closed rigid vessel such its temperature changes from 27 C to 427 C . Find the heat transfered
and change in internal energy .
Assume : -
R= 0.287 KJ/kg k
Cv = 0786 KJ/Kg k

Homework Equations



Q = mXCX(T2 - T1 )

The Attempt at a Solution



please can anyone explain to me how I can solve this question

and is R and Cv must be in same unit

my answer :

heat transfer = 1 X 0.786 X 10 (700-300 )

314.4X 10^3 KJ ( is the unit correct )

and about second one I don't have any idea for solve it
 
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ΔU=q + w(external)
w=work
as you posted Cv then volume is constnt so change in pressure should be taken.so by Gay Lussacs law P=kT
thereforeP1/P2=T1/T2
so you can find Work by volume*change in pressure
add Q to get work
 
I don't understand clearly can please help ?
 
Since the container is rigid, there is no volume change and no work is done. The heat transferred is equal to the change in internal energy.
 
thanks so much
 

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