Page 184 - Chemistry ICSE Class IX

172 ICSE Chemistry – 9
Solution: Given, Pressure = Constant
V = 500 cm 3
1
V = 260 cm 3
2
T = 373 K
1
T = ?
2
Using the Charles’ law equation,
V V
1 = 2
T 1 T 2
3
V × T 260 cm × 373 K
T = 2 1 = = 194 K
2
V 1 500 cm 3
EXAMPLE 7.6. A certain amount of a gas at 27°C and 1 atm pressure occupies
3
a volume of 25 m . If the pressure is kept constant and the temperature is raised
to 77°C, what would be the volume of the gas?
Solution: Here, pressure remains constant and temperature is changed, therefore,
from the Charles’ law,
V V
1 = 2
T 1 T 2
Now, V = 25 m 3
1
V = ?
2
T = 27 + 273.15 = 300.15 K
1
T = 77 + 273.15 = 350.15 K
2
Substituting these values in the above equation,
25 m 3 V
= 2
300.15 K 350.15 K
350.15 K
3
or V = × 25 m = 29.16 m 3
2
300.15 K
3
The gas would occupy a volume of 29.16 m .
EXAMPLE 7.7. Certain amount of a gas occupies a volume of 400 mL at 17°C.
To what temperature should it be heated so that its volume gets (i) doubled
(ii) reduced to half?
Solution:
(i) V = 400 mL
1
V = 2 × V = 800 mL
1
2
T = 17°C = (17 + 273.15) K = 290.15 K
1
T = ?
2
From the Charles’ law
V 1 = V 2
T 1 T 2
V
or T = 2 × T 1
2
V 1
Substituting the values of V , V and T , one gets
1
2
1
800 mL × 290.15 K
T = = 580.3 K
2
400 mL
or t = (580.3 – 273.15)°C = 307.15°C
2