A.13 process

$w=-P\Delta V$
and
$w=-\int_{V_{1}}^{V_{2}}PdV=-nRTln\frac{V_{2}}{V_{1}}$

Jessa Maheras 4F
Posts: 121
Joined: Fri Aug 02, 2019 12:16 am

A.13 process

For A.13,

A constant-volume calorimeter was calibrated by carrying out a reaction known to release 3.50 kJ of heat in 0.200 L of solu- tion in the calorimeter (q 5 23.50 kJ), resulting in a temperature rise of 7.32 8C. In a subsequent experiment, 100.0 mL of 0.200 m HBr(aq) and 100.0 mL of 0.200 m KOH(aq) were mixed in the same calorimeter and the temperature rose by 2.49 8C. What is the change in the internal energy of the reaction mixture as a result of the neutralization reaction?

The solution says to calibrate the calorimeter and then use q=CdeltaT, which gives deltaU and answers the question.

My question is, why don't we have to use the other info provided to ask the question? Do we not need to take moles or volume into account? Why not?

Justin Sarquiz 2F
Posts: 106
Joined: Fri Aug 30, 2019 12:15 am

Re: A.13 process

We do not take into account the moles of the strong acid and strong base because they are at the same molarity. Since they are a strong acid and a strong base, we assume that they combine to water, so we can just ignore that information in the problem.