### (a) Why is the change in enthalpy usually easier …

03:09
Manhattan College

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Problem 32

How much work (in J) is involved in a chemical reaction if the volume decreases from 5.00 to 1.26 L against a constant pressure of 0.857 atm?

$-325.17 \mathrm{J}$

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## Video Transcript

All right, guys. So we're doing Question three two of Cap Chapter five. We want to know how much work is involved off in this chemical reaction, where five five leaders of a container is compressed into one point six six leaders at a constant pressure of point eight five seven at GM. So we're going to use the standard war formula. Work is equal to negative pressure times the change in volume, so work is going to be equal minus zero point eight five seven p. M. Sometimes the change volume, which is going to be a V two minute. Anyone so one point two one point two six leaders Linus five leaders. And the answer to that is going to be B W is equal to three point two one leaders per ATM. Now they want us to find how much work is being done in jewels, so we have to use this conversion factor. One leader at GM physical to one's from one point thirty three jewels. It's a three point two one leader, times eighty eight and times one oh, one point three three jewels per one leader. It's AM and B that is going to give us three hundred and twenty five. I've point two six. Jules, This is how much work is being done on in our system. Three hundred twenty five point two six Jules.