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### What the calorimeter allows you to calculate according to conditions

Posted: Fri Feb 07, 2020 6:48 pm
In 4D, the textbook says:

A constant-pressure calorimeter and a constant-volume calorimeter measure changes in different state functions: at constant volume, the heat transfer is interpreted as DeltaU; at constant pressure, it is interpreted as DeltaH (Topic 4C).

Why is this so? From my understanding, it seems like this should be the other way around. Can someone explain?

### Re: What the calorimeter allows you to calculate according to conditions

Posted: Sat Feb 08, 2020 12:37 pm
At a constant volume, deltaV is zero, so there is no expansion work occurring (w=0) and deltaU overall will only equal q. At a constant pressure, since there may be expansion work occurring, the heat given off can only be interpreted as deltaH or q, since the overall internal energy includes the energy released/absorbed through expansion work done on/by the system as well as heat released/absorbed. Hope this helps!

### Re: What the calorimeter allows you to calculate according to conditions

Posted: Sat Feb 08, 2020 12:43 pm
Since deltaU = deltaH + -p*deltaV we can see that at deltaH = deltaU + p*deltaV. At constant volume, there is no change in volume so deltaV would equal 0 and at constant pressure, the equation still stands since the p value is just a constant anyways.

### Re: What the calorimeter allows you to calculate according to conditions

Posted: Sun Feb 09, 2020 2:56 pm
at constant volume, deltaV = 0, so there is no w expansion work being done, which means deltaU is only equal to q (deltaU=q+w). At constant pressure, expansion work may be occurring, and this change can only be deltaH or q as a result of the heat given off.

### Re: What the calorimeter allows you to calculate according to conditions

Posted: Wed Feb 12, 2020 9:48 am
Ariel Davydov 1C wrote:At a constant volume, deltaV is zero, so there is no expansion work occurring (w=0) and deltaU overall will only equal q. At a constant pressure, since there may be expansion work occurring, the heat given off can only be interpreted as deltaH or q, since the overall internal energy includes the energy released/absorbed through expansion work done on/by the system as well as heat released/absorbed. Hope this helps!

Thank you Ariel for the second time! That makes a lot of sense.... I feel like I want to meet you in person now since we've talked so much on here haha.

### Re: What the calorimeter allows you to calculate according to conditions

Posted: Sat Feb 22, 2020 7:46 pm
Jessa Maheras 4F wrote:
Ariel Davydov 1C wrote:At a constant volume, deltaV is zero, so there is no expansion work occurring (w=0) and deltaU overall will only equal q. At a constant pressure, since there may be expansion work occurring, the heat given off can only be interpreted as deltaH or q, since the overall internal energy includes the energy released/absorbed through expansion work done on/by the system as well as heat released/absorbed. Hope this helps!

Thank you Ariel for the second time! That makes a lot of sense.... I feel like I want to meet you in person now since we've talked so much on here haha.

Omg I would love to!