Energy with inflation of Balloon problem

[denverhockeyfan]

Homework Statement

A balloon is being inflated to its full extent by heating the air inside it. In the final stages of this process, the volume of the balloon changes from 4.00 x 106 L to 4.50 x 106 L by the addition of 1.3 x 108 J of energy as heat. Assuming that the balloon expands against a constant pressure of 1.0 atm, calculate delta E for the process.

Homework Equations

delta E = Efinal - Einitial
delta E = q + w
w = F x d

Pressure (P), Volume (V), Internal energy (E) and Enthalpy (H)
delta E = delta H +(- PdeltaV)

 

[GCT]

Homework Statement

A balloon is being inflated to its full extent by heating the air inside it. In the final stages of this process, the volume of the balloon changes from 4.00 x 106 L to 4.50 x 106 L by the addition of 1.3 x 108 J of energy as heat. Assuming that the balloon expands against a constant pressure of 1.0 atm, calculate delta E for the process.

Homework Equations

delta E = Efinal - Einitial
delta E = q + w
w = F x d

Pressure (P), Volume (V), Internal energy (E) and Enthalpy (H)
delta E = delta H +(- PdeltaV)

You need to display your work ... remember W = - P(deltaV).

 

[Blueshift5]

The first step in solving this problem is to calculate the work done by the balloon as it expands. This can be done using the equation w = F x d, where F is the force exerted by the balloon and d is the distance it expands. In this case, we can assume that the force is equal to the pressure, P, and the distance is the change in volume, delta V. Therefore, we can write w = P x delta V.

Next, we need to calculate the change in internal energy, delta E, for the process. This can be done using the equation delta E = q + w, where q is the heat added to the system and w is the work done by the system. In this case, we are given q = 1.3 x 108 J and we just calculated w.

Finally, we can use the relationship between enthalpy and internal energy, delta E = delta H + (-PdeltaV), to solve for delta H. This equation takes into account any changes in pressure during the process. In this case, the pressure is constant at 1.0 atm, so delta H = delta E + PdeltaV.

Substituting the values we have calculated, we get delta H = (1.3 x 108 J) + (1.0 atm x (4.50 x 106 L - 4.00 x 106 L)) = 1.8 x 108 J.

Therefore, the change in internal energy, delta E, for the process is 1.8 x 108 J. This means that the energy input to the system was converted into both work and an increase in internal energy.