For a liquid,the normal boiling point is -173 | vedclass

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(C) The Clausius-Clapeyron equation is given by: $\ln \frac{P_{2}}{P_{1}} = \frac{\Delta H_{vap}}{R} \left( \frac{1}{T_{1}} - \frac{1}{T_{2}} \right)$

Vedclass · Accepted Answer

$60$

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(C) The Clausius-Clapeyron equation is given by: $\ln \frac{P_{2}}{P_{1}} = \frac{\Delta H_{vap}}{R} \left( \frac{1}{T_{1}} - \frac{1}{T_{2}} \right)$Given: $P_{1} = 1\ atm$,$P_{2} = 2\ atm$,$T_{1} = -173 + 273 = 100\ K$,$\Delta H_{vap} = 200\ cal/mol$,$R = 2\ cal/mol\cdot K$,$\ln 2 = 0.7$.Substituting the values:$0.7 = \frac{200}{2} \left( \frac{1}{100} - \frac{1}{T_{2}} \right)$$0.7 = 100 \left( 0.01 - \frac{1}{T_{2}} \right)$$0.007 = 0.01 - \frac{1}{T_{2}}$$\frac{1}{T_{2}} = 0.01 - 0.007 = 0.003$$T_{2} = \frac{1}{0.003} = 333.33\ K$Converting to Celsius: $T_{2} = 333.33 - 273 = 60.33\ ^{\circ}C$.

For a liquid,the normal boiling point is $-173\,^{\circ}C$. At $2\ atm$ pressure,its boiling point should be nearly ......... $^{\circ}C$. (Given: $\Delta H_{vap} = 200\ cal/mol$,$R = 2\ cal/mol\cdot K$,$\ln 2 = 0.7$)

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