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

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

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

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?

Ariel Davydov 1C
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### Re: What the calorimeter allows you to calculate according to conditions

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!

ThomasNguyen_Dis1H
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Joined: Fri Aug 30, 2019 12:17 am

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

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.

Jesse H 2L
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### Re: What the calorimeter allows you to calculate according to conditions

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.

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

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

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.

Ariel Davydov 1C
Posts: 110
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### Re: What the calorimeter allows you to calculate according to conditions

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!