ecuaco

- A partir del dato anterior y de la ecuación de Clausius-Clapeyron calcule las diferentes presiones (PT) la temperatura correspondiente a cada presión y compárela con la obtenidaexperimentalmente. Complete la siguiente tabla.

Considere: Hv = 9700 cal / mol R = 1.987 cal / mol oK

Temperatura Experimental h2 h1 h = h2 - h1
P = h PT = 585 +PTemperatura Calculada
oC oK mmHg. mmHg. mmHg. mmHg. oC oK
22 oC 295 oK 9.0 mmHg 9.0 mmHg 0.0 mmHg 585 mmHg 22 oC 295 oK
96 oC 369 oK 10.0 mmHg 8.0 mmHg 2.0 mmHg 587 mmHg 95.9 oC 368.9 oK
98oC 371 oK 13.0 mmHg 5.0 mmHg 8.0 mmHg 593 mmHg 97.6oC 370.6 oK
100 oC 373 oK 16.0 mmHg 2.0mmHg 14.0 mmHg 599 mmHg 99.3 oC 372.3 oK

CALCULOS

Ecuación de Clausius-Clapeyron

lnP_2/P_1 = 〖∆H〗_V/R [1/T_1 - 1/T_2 ]

 T_2=1/(R[ln⁡〖P_2-ln⁡〖P_1 〗 〗/〖∆H〗_V ]+ 1/T_1 )

Variación de niveles de mercurio en el manómetro (h = h2 - h1)

h1 = 9.0 mmHg - 9.0 mmHg= 0.0 mmHg

h2 = 10.0 mmHg - 8.0 mmHg = 2.0 mmHg

h3 = 13.0 mmHg - 5.0 mmHg = 8.0 mmHg

h4 = 16.0 mmHg - 2.0 mmHg = 14.0 mmHg

Presión total (PT = 585 mmHg +P)

PT = 585mmHg + 0.0 mmHg = 585 mmHg

PT = 585 mmHg + 2.0 mmHg = 587 mmHg

PT = 585 mmHg + 8.0 mmHg = 593 mmHg

PT = 585 mmHg + 14.0 mmHg = 599 mmHg

Temperaturas calculadasT_2=1/(1.987 cal/(mol^o K) [ln⁡〖585 mmHg-ln⁡〖585 mmHg〗 〗/(9700 cal/mol)]+ 1/(〖295〗^o K))

T2 = 295 oK

T_2=1/(1.987 cal/(mol^o K) [ln⁡〖587 mmHg-ln⁡〖585 mmHg〗 〗/(9700 cal/mol)]+ 1/(〖369〗^oK))

T2 = 368.90 oK

T_2=1/(1.987 cal/(mol^o K) [ln⁡〖593 mmHg-ln⁡〖585 mmHg〗 〗/(9700 cal/mol)]+ 1/(〖371〗^o K))

T2 = 370.61 oK

T_2=1/(1.987 cal/(mol^o K) [ln⁡〖599 mmHg-ln⁡〖585mmHg〗 〗/(9700 cal/mol)]+ 1/(〖373〗^o K))

T2 = 372.32 oK

Temperatura de oK a oC
295 oK – 273 = 22 oC
368.9 oK - 273 =95.9 oC
370.6 oK - 273 = 97.6 oC
372.3 oK - 273 = 99.3 oC