How Do Delta U and Delta H Compare in a Constant-Volume Calorimetry?

[zmike]

Homework Statement

1 mole of a gas is burned completely burned in a constant-volume bomb calorimeter at 298 K, 5150 kJ of heat is evolved. Calculate the values delta U and delta H

Homework Equations

H = U + W

The Attempt at a Solution

I was lead to believe that you had to calculate number of moles difference delta n by my professor but that lead to no where.

Since constant volume, W = 0?? or does it?

if so then U would be = to qp then shouldn't q=5150 and U = H?

help??

 

[The legend]

I think...
yes ... you are right that the W=0 and dU=dH but "q=-5150 kJ" since heat is evolved..

so,

dU = q-W
so dU = q

looks like your work's done!

 

[Borek]

looks like your work's done!

If W=0, no work is done :tongue2:

 

[The legend]

If W=0, no work is done :tongue2:

:rofl:

 

[DeeNos]

First, it is important to clarify that the term "confusing thermodynamic q" is not appropriate. Thermodynamics is a complex and fundamental field of science, and it is important to approach it with patience and a willingness to learn. With that in mind, let's break down the problem at hand.

The problem states that 1 mole of gas is burned completely in a constant-volume bomb calorimeter at 298 K, resulting in the evolution of 5150 kJ of heat. This means that the reaction is exothermic, and the heat released is equal to the change in enthalpy (delta H) of the reaction. The value of delta H can be calculated by dividing the heat released by the number of moles of gas (1 mole).

However, the problem also asks for the values of delta U and delta H. These values are related by the equation H = U + W, where H represents enthalpy, U represents internal energy, and W represents work. In this case, since the reaction takes place in a constant-volume bomb calorimeter, there is no work being done (W = 0), so the equation simplifies to H = U. This means that the value of delta H is also equal to the change in internal energy (delta U) of the reaction.

To summarize, the values of delta U and delta H for this reaction are both equal to 5150 kJ. It is important to understand the concepts of enthalpy and internal energy, as well as the relationship between them, in order to solve thermodynamic problems effectively. I suggest reviewing these concepts and practicing with similar problems to strengthen your understanding.