thermodynamics - Isochoric process

[ahmedmoon]

¨A certain Gas of volume 0.4 m3, pressure of 4.5 bar
and temperature of 1300 C is heated to in a cylinder to 9 bar when the volume remains
constant. Calculate

 

¨ (i) Temperature at the
end of the process,

 

¨ (ii) the heat transferred

 

¨(iii) change in internal energy

 

¨ (iv) work done

 

¨(v) change in enthalpy.

 

¨Assume Cp = 1.005 kJ/kg.K and Cv=
0.71005 kJ/kg.K

my answer

p1v1=mRT

4.5 X 100 X 0.4 = m X 0.28 X 403

m= 1595 kg

now

i) p2v2 = mRT

9 X 100 X 0.4 = 1.1595 X 0.28 X T

T = 806 K

ii ) the heat transferred

no heat transferred because the work = 0

¨(iii) change in internal energy

Cv(T2 - T1 )

0.71005(806 - 40)

= 54.86

¨ (iv) work done = 0

¨(v) change in enthalpy.

Cp (T2 - T1 )

1.005(806 - 40)

=769.83

[daniton]

so, what is your question?

[ahmedmoon]

Hi

 

already I posted my question and my answer could please check my answer for each part

[daniton]

why do you say there is no heat transfer because W=0 that is not true work=0 means the whole heat is transferred to internal energy

the Change in ethalpy is equal to the Change in internal energy since work is zero.

[ahmedmoon]

the heat transferred = Cv(T2 - T1 )

0.71005(806 - 40)

= 54.86

[daniton]

what about the ethalpy change?

[ahmedmoon]

¨(v) change in enthalpy.

Cp (T2 - T1 )

1.005(806 - 40)

=769.83

[daniton]

why do you us Cp

since ethalpy =Internal energy + work done

[timo]

Since pV=NkT, for a process in which N and V (and k, obviously) are constant doubling the pressure should be equivalent to doubling the temperature. 806 Kelvin is not the double of 1300 °C (1573 K). So something went wrong. Note: I assumed a constant number of molecules, N, even though it is not explicitly stated. It seems natural that a container of gas heated from 4.5 bar to 9 bar (both in thermal equilibrium) does not permit particle exchange with the environment.

 

 

As a remark: I do refuse to even consider checking calculations that do not explicitly state the proper physical units of the terms being used. No offense meant by that, and especially no personal attack (I apply this rule to everyone): But from experience, many questions would not even exist if the person asking the question had been bothered adding units. And I refuse to invest my time under the condition that the people actually asking for help cannot be bothered to invest theirs.

[ahmedmoon]

Hi

Here we have ischoric process because the volume is constant

Now I will solve the question agine

 

I Found T2 = 806

 

By using p2/p1 = T2/T1

 

The heat transferred

Q = cv X delta T

0.71005(806 -403 )

 

= 228kj

 

Is Correct ..

Edited May 29, 2013 by ahmedmoon

[timo]

Well, it's rather obvious you used 130 °C rather than 1300 °C starting temperature. Be it an mistake in the calculation or a typo in your original post - who knows? I was actually serious about the "do not drop the physical" units thing. It's one of the most basic rules for doing physics. And one of the most important for doing it properly.