Page 180 - Chemistry ICSE Class IX

168 ICSE Chemistry – 9
From the data given here, we have

3
3
P (atm) V (dm ) PV (atm.dm )
2.00 11.2 22.4
1.00 22.4 22.4
0.90 24.7 22.2
0.75 29.9 22.4
0.50 44.4 22.2
Average value = 22.3 ± 0.1
Since the value of PV is nearly constant, the given data verify the Boyle’s law.
3
EXAMPLE 7.2. 100 cm of a gas enclosed in a vessel maintained at a pressure
3
of 1400 Torr, is allowed to expand to 250 cm under constant temperature
conditions. What would be its pressure?
Solution: Since, temperature remains constant, hence
1
V v or PV = Constant
P

Therefore, P V = P V 2
1
2
1
P = 1400 Torr V = 100 cm 3
1
1
P = ? V = 250 cm 3
2
2
P V 1400 × 100
This gives, P = 1 1 = = 560 Torr
2
V 2 250
The nal pressure of the gas after expansion would be 560 Torr.
EXAMPLE 7.3. A student while measuring pressure and volume of a gas at
constant temperature forgot to record some of the observations as follows.
Pressure (Torr) 100 125 200 (–)
Volume (mL) 80 (–) 40 32
Calculate the missing observations. Name the law used in your calculations.
Solution: In this experiment, temperature is kept constant and effect of pressure
on volume of a certain amount of gas is measured. Therefore, the data can be
analysed in terms of the Boyle’s law, i.e.,
P V = P V = P V = P V 4
3
4
3
2
2
1
1
Now, from rst observation,
P = 100 Torr and V = 80 mL
1
1
.
.
Thus, P V = 100 × 80 Torr mL = 8000 Torr mL
1
1
From second observation,
P = 125 Torr and V = ?
2
2
.
Therefore, P V = P V = 8000 Torr mL
1
1
2
2
.
8000 Torr mL
or V = = 64 mL
2
125 Torr
Thus, the missing volume in the second observation is 64 mL. Similarly, for the
fourth reading, we can write,
.
P × 32 mL = 8000 Torr mL
4
.
8000 Torr mL
Hence, P = = 250 Torr
4
32 mL
Thus, the missing pressure in the fourth observation is 250 Torr.