# Isobaric process

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$$\Delta$$V = nR$$\Delta$$T

w = nR$$\Delta$$T

## The Attempt at a Solution

w = 610 J; n = 3.4

610/(3.4*8.314) = 21.5 K = $$\Delta$$T

Calculations look right to me,but the answer is 8.63 K

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Andrew Mason
Homework Helper
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$$\Delta$$V = nR$$\Delta$$T

w = nR$$\Delta$$T

## The Attempt at a Solution

w = 610 J; n = 3.4

610/(3.4*8.314) = 21.5 K = $$\Delta$$T

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

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.