4C.3


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Ami_Pant_4G
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Joined: Sat Aug 24, 2019 12:17 am

4C.3

Postby Ami_Pant_4G » Mon Feb 10, 2020 2:28 pm

Calculate the final temperature and the change in enthalpy when 765 J of energy is transferred as heat to 0.820 mol Kr(g) at 298 K and 1.00 atm (a) at constant pressure; (b) at constant volume. Treat the gas as ideal.

Can someone please explain how to find the final temperature for both parts? Also how would I find enthalpy for part b? Thanks in advance.

Bella Townsend
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Joined: Wed Feb 20, 2019 12:18 am

Re: 4C.3

Postby Bella Townsend » Mon Feb 10, 2020 2:44 pm

Some equations you need to know for this equation is when you are calculating the Cpm for a molecule of one type of atom like an ideal gas you use q=n*5/2R*deltaT
when you are calculating the Cvm for a molecule of one type of atom like an ideal gas you use q=n*3/2R*deltaT

The rest is given to you

765=0.820 (5/2 * 8.314) (Tf - 298)
Tf=343 K

At constant pressure the enthalpy is equal to the heat so delta h= 765J

765=0.820 (3/2 * 8.314) (Tf - 298)
Tf=373 K

MAC 4G
Posts: 121
Joined: Wed Sep 18, 2019 12:16 am

Re: 4C.3

Postby MAC 4G » Mon Feb 10, 2020 3:05 pm

The above post is correct in how you find delta T, I found that it was easier to delta T = q/nCv,m and then use the equation Tf = delta T + Ti. Then to find enthalpy of part b, you would use the equation delta H = q + nRdeltaT and plug in all the values you were given and the delta T value you found in part a (assuming you did it the way that I did solving for delta T before Tf)

Matthew ILG 1L
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Joined: Sat Aug 17, 2019 12:15 am

Re: 4C.3

Postby Matthew ILG 1L » Mon Feb 10, 2020 4:37 pm

Is the change in enthalpy supposed to be much larger for part b than it is for part a?

MAC 4G
Posts: 121
Joined: Wed Sep 18, 2019 12:16 am

Re: 4C.3

Postby MAC 4G » Mon Feb 10, 2020 4:41 pm

Matthew ILG 1L wrote:Is the change in enthalpy supposed to be much larger for part b than it is for part a?


Depends on what you define as much larger, assuming I calculated both correctly, part b is 510 J more than part a's enthalpy.

DHavo_1E
Posts: 118
Joined: Sat Aug 17, 2019 12:17 am

Re: 4C.3

Postby DHavo_1E » Mon Feb 10, 2020 11:31 pm

Hello,

Can I ask why for constant volume we do not get an answer for the change in enthalpy? Thank you

Matthew Chan 1B
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Joined: Sat Sep 07, 2019 12:16 am

Re: 4C.3

Postby Matthew Chan 1B » Tue Feb 11, 2020 2:01 am

DHavo_1E wrote:Hello,

Can I ask why for constant volume we do not get an answer for the change in enthalpy? Thank you


This is because at a constant volume, qv = of the system.

DHavo_1E
Posts: 118
Joined: Sat Aug 17, 2019 12:17 am

Re: 4C.3

Postby DHavo_1E » Tue Feb 11, 2020 12:25 pm

Matthew Chan 1B wrote:
DHavo_1E wrote:Hello,

Can I ask why for constant volume we do not get an answer for the change in enthalpy? Thank you


This is because at a constant volume, qv = of the system.



Hello,

Can I ask that when you found qv=deltaV whether you used the equation q=U-w? Is the enthalpy equation: DeltaH=DeltaU+pDeltaV used only for constant pressure but not constant volume? Thank you!!


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