Thermodynamics: Stiff and isolated container

[Karol]

Homework Statement

A stiff and isolated vessel contains oxygen and hydrogen. a spark is made and explosion occurs. temperature and pressure rise high. apply the first law.

Homework Equations

The first law: $Q=U_2-U_1+W$

The Attempt at a Solution

Is the heat generated in the explosion considered incoming heat? if not, U₂=U₁ but it's not logical, how can the final state, with it's higher pressure and temperature be equal to the initial? can the internal energy be the same?

 

[ehild]

Homework Statement

A stiff and isolated vessel contains oxygen and hydrogen. a spark is made and explosion occurs. temperature and pressure rise high. apply the first law.

Homework Equations

The first law: $Q=U_2-U_1+W$

The Attempt at a Solution

Is the heat generated in the explosion considered incoming heat?

No, it is change of the chemical energy of the constituents. If chemical reaction can occur in a system, you have to include chemical energy to the internal energy.

 

[Chestermiller]

Homework Statement

A stiff and isolated vessel contains oxygen and hydrogen. a spark is made and explosion occurs. temperature and pressure rise high. apply the first law.

Homework Equations

The first law: $Q=U_2-U_1+W$

The Attempt at a Solution

Is the heat generated in the explosion considered incoming heat? if not, U₂=U₁ but it's not logical, how can the final state, with it's higher pressure and temperature be equal to the initial? can the internal energy be the same?

Yes, you are correct. The internal energy is the same. But, how can this be?

What would you have to do if the reactor were not isolated, and you wanted the final temperature of the water to be the same as the initial temperature of the hydrogen and oxygen (assuming in the final state, the water is still a vapor)? What does that tell you about the internal energy of the water (final state) at that constant temperature compared to the internal energy of the reactants (initial state)?

Chet

 

[Karol]

What would you have to do if the reactor were not isolated, and you wanted the final temperature of the water to be the same as the initial temperature of the hydrogen and oxygen

Maybe i don't understand but if i want to cool the vapor i would take out heat from the reactor, and it shows that the internal energy of the hot vapor in the final state is higher than at the beginning because temperature is a measure of internal energy, so the less i understand

 

[Chestermiller]

Maybe i don't understand but if i want to cool the vapor i would take out heat from the reactor, and it shows that the internal energy of the hot vapor in the final state is higher than at the beginning because temperature is a measure of internal energy, so the less i understand

You almost had it. If, to maintain the temperature constant, you need to remove heat, the internal energy of the products relative to the reactants must be negative. We call such a reaction "exothermic" and we say that the "heat of reaction" (the internal energy change at constant temperature) is negative. That is, in order for Q to be negative, ΔU must be negative to maintain constant temperature. Now if, in the isolated system, ΔU = 0, that means that the temperature of the water vapor must rise above the original temperature of the hydrogen and oxygen.

Chet