Hi Geniuses,

Just need some help to check whether i have used the right calculations to work out a problem.

the question states:

Calculate the

**final temperature ** of a sample of carbon dioxide (assume ideal gas) of

**mass 16g ** that is expanded

**reversibly and adiabatically** from

**500mL to 2.00L**. The constant pressure heat capacity of CO

_{2} is

**37.11 J k**^{-1} mol^{-1} .

Here is my working:

1.) the process is adiabatic and reversible, therefore,

**(heat)q=0, **

(work) w=0 and internal energy change

**delta U=0**.

2.) entropy change,

**delta S = q**_{rev}/T= 0So, if the pressure of the system is held constant (delta P=0) as indicated by the constant pressure heat capacity value given in the question and the moles (n) are also held constant, then the total entropy change of the system can be calculted by using the following equation:

delta S

_{sys}= n*C

_{p}ln(T

_{final} / T

_{initial}) + n*R ln(V

_{final}/V

_{initial})

Now if delta S is zero for a reversible adiabatic process, the above equation is equal to zero, therefore, the final temperature can be calculated by solving this equation. ( can't it

)

Looking at the values given in the question:

n= 16g*(1mol CO

_{2} / 44.0098g) = 0.36355589mol CO

_{2}V

_{initial}=0.5L

V

_{final}=2.0L

T

_{initial}=298.15K

C

_{p}=37.11 J K

^{-1} mol

^{-1}and inserting into the equation:

[(0.36355589mol)(37.11 J K

^{-1} mol

^{-1}) ln (T

_{final} / 298.15K)] + [(0.36355589mol)(8.314472 J K mol) ln (2.0L / 0.5L)] = 0

Which simplifies to:

(13.49155908 J K

^{-1}) ln (T

_{final} / 298.15K) + 4.190456309 J K = 0

so:

(13.49155908 J K

^{-1}) ln (T

_{final} / 298.15K) = -4.190456309 J K

Rearranging:

ln (T

_{final} / 298.15K) = -4.190456309 J K / 13.49155908 J K

^{-1}so:

ln (T

_{final} / 298.15K) = -0.310598373

then:

(T

_{final} / 298.15K) = e

^{-0.310598373}And finally:

T

_{final} = (e

^{-0.310598373})*(298.15K)

= 218.5463986K

Rounding off:

T

_{final} = 218.55K

If anyone can confirm that this was the right method to use i would appreciate it greatly, is there a simpler way to calculate final temperature in a reversible adiabatic process?

cheers,

madscientist :albert: