Question

A student forgot to add the reaction mixture to the round bottomed flask at 27°C but instead he/she placed the flask on the flame. After a lapse of time, he realized his mistake, and using a pyrometer he found the temperature of the flask was 477°C. What fraction of air would have been expelled out?

Answer 1

Consider the volume of the flask = $V cm^3$

Let this be $V_1$ at 27°C

$V_2$ = ?

Given data:

$T_1$ = 27°C = 27 + 273 = 300K

$T_2$ = 477°C = 477 + 273 = 750K

According to Charles law,

$V_1/T_1 = V_2/T_2$

Substitute the values.

$V_2 = \frac{(V\times 750)}{300}$

$V_2 = 2.5 V$

The volume remaining in the flask is V

Volume of expelled air = (2.5-1)V = 1.5 V

The fraction of expelled air = 1.5V/2.5V = $\frac{3}{5}$

Answer 2

Answer:

iiiThe initial temperature T=27

o

C=27+273K=300K

The final temperature T

′

=477

o

C=477+273K=750K

The inital volume is V L and the final volume is V' L.

V

′

V

=

T

′

T

V

′

V

=

750

300

V

′

V

=

5

′

2

The fraction of air expelled is 1−

V

′

V

=1−

5

2

=

5

3

.

Explanation:

mar