KPK 11th Class Chemistry Chapter 4 States of Matter-I Gases Short Questions Answers
KPK 11th Class Chemistry Chapter 4 States of Matter-I Gases Short Questions with answers are combined for all 11th class(Intermediate/hssc) Level students.Here You can prepare all Chemistry Chapter 4 States of Matter-I Gases short question in unique way and also attempt quiz related to this chapter. Just Click on Short Question and below Answer automatically shown. After each question you can give like/dislike to tell other students how its useful for each.
Class/Subject: 11th Class Chemistry
Chapter Name: States of Matter-I Gases
Board: All KPK Boards
- Malakand Board 11th Class Chemistry Chapter 4 States of Matter-I Gases short questions Answer
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KPK 11th Class Chemistry Chapter 4 States of Matter-I Gases Short Questions Answers
Since the number of moles any gas is related directly to its volume i.e V ∝ n.
There force, it is expected that one mole of all gases occupy the same volume at a given temperature and pressure. So atr S.T.P one mole of any das occupies 22.4 dm3 e.g.
22.4dm3 of H2 at S.T.P = 22.4dm3
1cm3 or 0.001dm3at S.T.P = 22.4 0.001=0.0224dm3
22.4dm3 of CH4 at S.T.P =22.4dm3
1cm2 or 0.001 dm3 at S.T.P =22.40.001=0.0224dm3
Hence the volumes of the two gases are equal.
There fore, they have equal number of molecules.
- The actual volume of the gas is negligible as compared to the volume occupied by the gas.
- No force of attraction or repulsion is present between the molecules and each molecule behaves independently. The first postulate is responsible for the deviation at high pressure and second postulate for the deviation at low temperature.
This is because at high pressure the gas molecules come closed, the empty space between the molecules decreases and as a result the volume occupied by the actual gas molecules does not remain negligible as compared to the total volume of the gas. The gas molecules don’t have attraction for each other at high temp. The K.E of the molecules is very large and the attractive force becomes negligible. However, when intermolecular attractive forces no longer remain negligible and the collision between the molecules no longer remain perfectly elastic.
It is observed that there are two assumptions in the K.M.T.
1. The actual volume of the gas it negligible as compared to the volume occupied by the gas.
2. No force of attraction or repulsion is present between the molecules and each molecule behaves independently. The first postulate is responsible for the deviation at high pressure and second postulate for the deviation at low temperature.
The compressed air is then passed through a spiral tube having a jet at the end. Where the gas comes out of jet into low pressure area (1atm) it is cooled down due to Joule Thomson effect. The cooled air moves up, cools the incoming gas of the jet and then enters into the compression pump to be compressed again. By repeating compression and expansion process again and again, air is liquefied. The apparatus is shown in the figure
Ammonia is seen to diffuse about 1.5 times as fast as HCL. Moreover since the rate of diffusion is proportional to the distance traveled by the gas thus we can write:
RNH3/rHCL=distance traveled by NH3/distance traveled by HCl = 1.51
According to the graham’s law of diffusion.
rA
rNH3/rHCL = √dHCL/dNH3
= √1.66/0.76 = 1.5
Similarly rNH3
rNH3/rHCL √MHCL/MNH3 MNH3=17a.m.u
MHCL = 36.5a.m.u
= 1.465 or 1.5
The above calculations verify Graham’s law of diffusion.
Example : Determine the relative rates of diffusion of equal volumes of H2 and CO2 under the same conditions of temperature and pressure.
