4A.3

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RRahimtoola1I
Posts: 102
Joined: Fri Aug 09, 2019 12:15 am

4A.3

Postby RRahimtoola1I » Wed Jan 29, 2020 10:29 pm

Can someone please explain this problem. The solution manual is using 101.325J as a conversion but I'm not sure where they are getting it from. Other than that, I still don't understand the problem.

rachelle1K
Posts: 109
Joined: Sat Sep 07, 2019 12:16 am

Re: 4A.3

Postby rachelle1K » Thu Jan 30, 2020 12:05 pm

The purpose of using 101.325 is to convert atm to Pa. In SI units, Pa is J.m^-3 . The equation used is w=PAD, and work is in joules, so the units would essential cancel out to joules because w=(J.m^3)(m^2)(m) .

Ellis Song 4I
Posts: 102
Joined: Thu Jul 11, 2019 12:17 am

Re: 4A.3

Postby Ellis Song 4I » Thu Jan 30, 2020 6:37 pm

You can find this number in the constants and equations paper. It converts 1 Latm to 101.325 J.

BritneyP- 2c
Posts: 101
Joined: Sat Sep 14, 2019 12:15 am

Re: 4A.3

Postby BritneyP- 2c » Thu Jan 30, 2020 6:47 pm

The textbook explains that the number comes from converting L*atm to Joules.


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