The Calculator Minimalist Programming Club

[Thomas Klemm]

(03-03-2019 11:55 AM)Csaba Tizedes Wrote:  One application on my TI-57 LCD - maybe it is enough for membership:
TI 57 LCD calculator solves Clausius Clapeyron with Euler method

For those interested here's the program for the TI-57II:

Code:

LBL 0   23.00     00
RCL 0   71.00     01
÷       55        02
RCL 1   71.01     03
x²      34        04
×       65        05
RCL 2   71.02     06
×       65        07
RCL 3   71.03     08
=       95        09
STO+ 0  61.85.00  10
RCL 3   71.03     11
STO+ 1  61.85.01  12
RCL 0   71.00     13
x<t    -27        14
GTO 0   22.00     15
Pause   96        16
RCL 1   71.01     17
–       75        18
RCL 3   71.03     19
÷       55        20
2       02        21
=       95        22
R/S     13        23

We can solve this differential equation by separating the variables:

\(\int_{p_0}^{p_b} \frac{dp}{p}=C \int_{T_0}^{T_b} \frac{dT}{T^2}\)

This leads to:

\(\log(p_b) - \log(p_0) = C \left (\frac{1}{T_0} - \frac{1}{T_b} \right )\)

Which allows us to solve for \(T_b\):

\(\begin{align*}
\frac{1}{T_b} &= \frac{1}{T_0} - \frac{\log(p_b) - \log(p_0)}{C} \\
&= \frac{1}{T_0} + \frac{\log(p_0) - \log(p_b)}{C} \\
&= \frac{1}{T_0} + \frac{\log(\frac{p_0}{p_b})}{C} \\
\end{align*}\)

Given the following values we can calculate the temperature \(t_b = T_b - 273.15\) of the boiling point:

\(\begin{matrix}
T_0 =&100 + 273.15 \\
p_0 =&101325 \\
p_b =&200000\\
C =& 4889
\end{matrix}\)

100 + 273.15 =
1/x + (101325 ÷ 200000) lnx ÷ 4889 =
1/x - 273.15 =
120.42637

Cheers
Thomas