Solving Isobaric Process Heat Addition Problem

AI Thread Summary
The discussion revolves around solving a heat addition problem for a monatomic ideal gas under isobaric conditions. Initially, the user calculated the change in temperature using the work equation but arrived at an incorrect result. It was clarified that the heat added contributes to both work done and an increase in internal energy, necessitating the use of the specific heat equation. The correct specific heat for a monatomic gas was identified as 5/2R, leading to the successful calculation of the temperature change. The user confirmed they resolved the issue after recognizing the mistake.
yankees26an
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NVM I solved it. Thanks for reading anyway :)

Homework Statement


610 J of heat is added to 3.4 mol of a monatomic ideal gas at constant pressure. Find the
change of temperature of the gas.

Homework Equations


At isobaric conditions, w = p\DeltaV = nR\DeltaT

w = nR\DeltaT

The Attempt at a Solution



w = 610 J; n = 3.4

610/(3.4*8.314) = 21.5 K = \DeltaT

Calculations look right to me,but the answer is 8.63 K
 
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yankees26an said:
NVM I solved it. Thanks for reading anyway :)

Homework Statement


610 J of heat is added to 3.4 mol of a monatomic ideal gas at constant pressure. Find the
change of temperature of the gas.


Homework Equations


At isobaric conditions, w = p\DeltaV = nR\DeltaT

w = nR\DeltaT


The Attempt at a Solution



w = 610 J; n = 3.4

610/(3.4*8.314) = 21.5 K = \DeltaT

Calculations look right to me,but the answer is 8.63 K
You are equating Q with W. That is true only if the internal energy does not change, which of course is not the case when temperature increases.

The 610 joules of heat flow does two things: causes the gas to do work and increases its internal energy.

The specific heat at constant pressure is the heat flow per degree of temperature rise (per mole) of a monatomic gas during a constant pressure process:

\Delta Q = nC_p\Delta T

Use that to calculate the change in temperature. What is Cp for a monatomic gas?

AM
 
Andrew Mason said:
You are equating Q with W. That is true only if the internal energy does not change, which of course is not the case when temperature increases.

\Delta Q = nC_p\Delta T

Use that to calculate the change in temperature. What is Cp for a monatomic gas?

AM

Yea I realized that and used the other equation after some trial and error. Cp for a monoatomic gas is 5/2R :)

Solved.
 
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