Question

For a liquid normal boiling point is -173°C, then at 2 atm pressure its boiling point should be nearly (∆H vap. = 200 cal/mol)
A)-73°C
B)333°C
C)60°C
D)103°C

Answer 1

Answer:

A) -73°C

is the answer

Explanation:

Hope you like the answer

Answer 2

The correct option is c) 60° C

Given:

T1 = -173°C

P1 = 1 atm ( as it is under normal conditions)

P2 = 2 atm

∆H vap. = 200 cal/mol

To Find:

The temperature of the given liquid at 2 atm pressure

Solution:

$\frac{P2}{P1} =$ ∆H vap/2 . $\frac{T2-T1}{T1T2}$

$In(2)=\frac{200}{2}(\frac{T2-T1}{T1T2} )$

So the final answer is

T2 = 333°K = 60°C

This is the temperature required at 2 atm pressure.