# Homework Help: Thermodynamics process

1. Jun 12, 2007

### blumfeld0

1. The problem statement, all variables and given/known data

i have a thermdyanmics process. with a constant pressure, constant volume and adiabatic. please see the attachment.

2. Relevant equations

from C->A (constant pressure)
so change in pressure = 0
equations needed to do this problem:

Pressure* V = n *R* T -----------> main equation
heat = Q = Cp * n * change in temperature ----------------> main
(Cp is called the specific heat at constant pressure)

3. The attempt at a solution

our goal is use Q = Cp n * change in temperature

BUT we don't know change in temperature
we know Cp = 20.8 for an ideal monatomic gas (just a constant you look up in a physics textbook)
n = 1 mole

so how do we find change in temperature?

we need to use P Vinitial = n R Tinitial
400 * (1) = 1 * 8.315 * Tinitial

so Tinitial = 48.1058 Kelvin

P Vfinal = n R Tfinal

400 * 2 = 1 * 8.315 * Tfinal
so Tfinal = 96.2117 Kelvin

now we are done!

Q = n * Cp * change in temperature

Q = (1) *( 20.8 J/mol*K ) (96.2117 - 48.1058 Kelvin)
Q = 1000.6 Joules

is this right?

thank you

#### Attached Files:

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2. Jun 12, 2007

### Andrew Mason

This is just a first law problem: dQ = dU + dW where dW is the work done by the gas. For C to A:

(1) $$\Delta Q_{CA} = \Delta U_{CA} + W_{CA} = \Delta U_{CA} + P\Delta V_{CA}$$

You know that the change internal energy from C to A is +100 J (to balance the 200 J loss from A to B and 100 J gain from B to C).

You can also determine the work done from C to A because you are given the pressure and change in volume.

So plug those values in (1) to give you $\Delta Q_{CA}$

AM

3. Jun 12, 2007

### blumfeld0

thank you!
why was i off? i thought i had this!

so Q = U + W
Q = 100 + (400pascals * (2-1)m^3)
Q= 500 joules?

or is it

Q = 100 - (400pascals * (2-1)m^3)
Q= -300 joules.

which one is right?

thank you!

4. Jun 12, 2007

### Andrew Mason

Heat flow out of the gas is negative. Heat flow into the gas is positive. Work done by the gas is positive. Work done on the gas is negative.

So:

$$\Delta Q = \Delta U + P\Delta V = +100 + 400 * (-1) = -300 J$$

This means that 300 J of heat flows out of the gas in going from C to A.

AM