The Calculator Minimalist Programming Club
03-03-2019, 04:27 PM
Post: #21
 Thomas Klemm Senior Member Posts: 1,885 Joined: Dec 2013
RE: The Calculator Minimalist Programming Club
(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:

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

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

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
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