Why do we change the use of the heat formula in the following problem from using Cp to then using Cv:

Problem: A gas is initially at 38 ̊C and having Cp= 0.92 KJ/kg.K and Cv=0.71KJ/kg.K is

placed within a cylinder. If 8.5 KJ of heat is added to 5 kg of the gas in a non-

work flow, constant pressure process. Determine:

(a) The final gas temperature

(b) Work done on or by the gas

Solution:

We use Q = mCp (T2 - T1) to find T2 because it is a constant pressure process; but then use

Q = mCv (T2 - T1) to find work.

Shouldn't we still use

Q = mCp (T2 - T1) to find Work ?

Thanks

The work they have calculated is the one that is consumed against atmosphere. Isn't it? They have assumed the process to be constant volume and found out the amount of heat that would have been required to raise the temp by 1.84 degrees Celcius. And then subtracted this heat transfer from previous one. This is not the right approach to calculate thermodynamic work done by the system. The work done by the system should have been difference between initial and final volumes multiplied by pressure. As we do not know either pressure or volume difference of piston we can not calculate work done by the system. Whatever they have calculated does not represent anything related to Thermodynamic work done by piston that can be useful.

So actually we can't find the work done as we don't know the change in volume?

𝑞=𝑄/𝑚 = ∆ℎ=ℎ2―ℎ1=𝑐𝑝(𝑇2―𝑇1)

I also got the same answer for T2 in part (a) but I used simple Q =mCpdT.

What is given in this solution seems to be a printing mistake.

Especially this line,

𝑚=∆ℎ=ℎ2―ℎ1=𝑐𝑝(𝑇2―𝑇1)

part (b) is wrong.

You can not use Cv because the process is done under constant pressure. At constant pressure, the amount of heat required to raise the temp of the unit amount of substance is greater than that which is required at constant volume because some amount of heat is consumed to raise the piston and to do the work against the atmosphere.

To find out work done in this process you need either pressure value or initial volume everything else can be calculated using these values.

I thin there's a typo somewhere; here is the given solution:

The process is carried out in constant volume conditions. Using the constant

volume specific heat equation, we get:

𝑞=𝑄

𝑚=∆ℎ=ℎ2―ℎ1=𝑐𝑝(𝑇2―𝑇1)

𝑞=8.5

5=0.92(𝑇2―𝑇1)

8.5

5=0.92(𝑇2―𝑇1)

1.7=0.92(𝑇2―311)

𝑇2=312 𝐾

(b) To obtain the work, we will use the first law equation for the closed

systems:

𝑸―𝑾=∆𝑼

―𝑾=∆𝑼―𝑸

―𝑾=𝒎𝑪𝒗(𝑇2―𝑇1)―𝑸

―𝑾=𝟓×𝟎.𝟕𝟏(1.84)―𝟖.𝟓

𝑾=𝟏.𝟗𝟔𝟖𝑲𝑱

To find out work done in a constant pressure process we use the following relation.

Q = mCvdT

gives the heat transferred into or out of the system during the constant volume process and

not work done.Interpretation of this process:In this constant pressure heat addition process usually, the work is done by the system as the piston moves away to keep the pressure constant. The movement of the piston results in the work done by the system and it is positive.