A) 250 K
B) 294 K
C) 51.6 K
D) 12.5 K
Correct Answer: B
Solution :
[b] Applying ClausiusClapeyron equation, we get |
\[\log \frac{{{P}_{2}}}{{{P}_{1}}}=\frac{\Delta {{H}_{v}}}{2.303R}\left[ \frac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}\times {{T}_{2}}} \right]\] |
\[\log \frac{760}{23}=\frac{40656}{2.303\times 8.314}\left[ \frac{373-{{T}_{1}}}{373{{T}_{2}}} \right]\] |
This gives \[{{T}_{1}}\] = 294.4 K. |
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