4B.5

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vanessas0123
Posts: 100
Joined: Wed Sep 11, 2019 12:17 am

4B.5

Postby vanessas0123 » Wed Feb 05, 2020 5:19 pm

An ideal gas in a cylinder was placed in a heater and gained
5.50 kJ of energy as heat. If the cylinder increased in volume from
345 mL to 1846 mL against an atmospheric pressure of 750. Torr
during this process, what is the change in internal energy of the
gas in the cylinder?

Which equation would we use for this?

Amanda Lin 2I
Posts: 101
Joined: Sat Aug 17, 2019 12:15 am

Re: 4B.5

Postby Amanda Lin 2I » Wed Feb 05, 2020 5:52 pm

You would use w=-PΔV, solve for w, then plug w into ΔU=q+w.

Shutong Hou_1F
Posts: 117
Joined: Sat Sep 14, 2019 12:17 am

Re: 4B.5

Postby Shutong Hou_1F » Sat Feb 08, 2020 12:23 am

Yes, and remember to pick ideal gas constant with correct units or convert 750 torr to 750/760 atm; also remember to be consistent with either J or kJ when adding energies.

Pegah Nasseri 1K
Posts: 100
Joined: Wed Feb 27, 2019 12:15 am

Re: 4B.5

Postby Pegah Nasseri 1K » Sat Feb 08, 2020 12:41 am

You would use the equation delta U = q + w. You know that the gas gained 5.50 kJ of heat energy so q will be positive (q = 5500J). To find the work, you must use the equation W = - (P*delta V). However make sure to convert the volumes given to you from milliliters to liters and to convert the pressure from Torr to atm (1 atm = 760 Torr). You must also convert the calculated value of your work done into the proper units because it will be calculated in the units L*atm and it needs to be joules. This is done by multiplying the value by 101.325 J (1 L*atm = 1 101.325J). You can then plug in your q value and your w value into the equation to determine the change in internal energy.

Matthew ILG 1L
Posts: 112
Joined: Sat Aug 17, 2019 12:15 am

Re: 4B.5

Postby Matthew ILG 1L » Sun Feb 09, 2020 11:09 pm

Will the conversion of 1atm=760 Torr be given to us at all? I have yet to find that on the equations sheet from test one. I could be looking in all the wrong places though.

Sofia Barker 2C
Posts: 101
Joined: Wed Sep 18, 2019 12:21 am

Re: 4B.5

Postby Sofia Barker 2C » Sun Feb 09, 2020 11:35 pm

You should use the equations delta U = q+w and w = -P(delta V). q is given as 5.50 kj, but w is not given. Since the external pressure is constant, use w = -P(delta V) with P in atm and V in liters so that you can use the conversion of 101.325 joules per L atm. Then convert q or w to joules or kilojoules and then sum the two values to get the change in internal energy, delta U.

205154661_Dis2J
Posts: 109
Joined: Wed Sep 18, 2019 12:21 am

Re: 4B.5

Postby 205154661_Dis2J » Sun Feb 09, 2020 11:49 pm

We know that internal energy is q+w. We know from the question that q is 5.50kJ. Now we need to figure out the work done. Thus, we would be using the equation w = -PΔV to figure that out. Then when work is calculated, we add 5.50kJ + work done and that will be the change in internal energy.

Kavya Immadisetty 2B
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Re: 4B.5

Postby Kavya Immadisetty 2B » Tue Feb 11, 2020 4:48 pm

Matthew ILG 1L wrote:Will the conversion of 1atm=760 Torr be given to us at all? I have yet to find that on the equations sheet from test one. I could be looking in all the wrong places though.


It's on the constants and formula sheet


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