8.31

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Angela 1K
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Joined: Fri Sep 29, 2017 7:05 am

8.31

Postby Angela 1K » Thu Jan 18, 2018 8:00 pm

This question explicitly asks to calculate the heat released. I was wondering how you know to find molar heat capacity (and thus you have to change the mass into number of moles, as shown in the solutions manual) rather than keeping the Kr in terms of grams.

Thanks.

Dylan Mai 1D
Posts: 70
Joined: Sat Jul 22, 2017 3:00 am

Re: 8.31

Postby Dylan Mai 1D » Thu Jan 18, 2018 8:12 pm

I believe you know because they say assume that krypton behaves as an ideal gas, and therefore you would have to use the constants 5/2R and 3/2R which are molar heat capacities

Tim Nguyen 2J
Posts: 31
Joined: Sat Jul 22, 2017 3:00 am

Re: 8.31

Postby Tim Nguyen 2J » Thu Jan 18, 2018 8:53 pm

As the response stated above iterates, since you know that Krypton is assumed to behave as an ideal gas, you must refer to the molar heat capacity constants for ideal gases under constant pressure and volume (in this case, 5/2R and 3/2R respectively. Using those values in q= n(molar heat capacity under constant volume OR pressure)(delta T), the amount of heat released may be calculated.

cristiancampana 2H
Posts: 32
Joined: Fri Sep 29, 2017 7:07 am

Re: 8.31

Postby cristiancampana 2H » Wed Jan 24, 2018 11:01 pm

in page 281, it says that Cv,m= 3/2R . so you multiply both sides by m to make Cv,m just Cv. Then Cv= ΔU/ΔT. multiply both sides by ΔT and you get ΔU= 3/2R n ΔT. and ΔU is just q because w is 0 since there is no change in volume.

Vincent Chiang 1L
Posts: 54
Joined: Fri Sep 29, 2017 7:05 am

Re: 8.31

Postby Vincent Chiang 1L » Thu Jan 25, 2018 10:44 pm

On the same question but regarding units, can molar heat capacity be J/(mol.C)? It thought it had to be J/(mol.K)...

Nishma Chakraborty 1J
Posts: 51
Joined: Fri Sep 29, 2017 7:04 am

Re: 8.31

Postby Nishma Chakraborty 1J » Thu Jan 25, 2018 10:55 pm

Molar heat capacity can be either J/(mol∙C) or J/ (mol∙K)


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