Figures below depict air enclosed in a cylinder by an air-tight piston. The piston has been pushed as shown in figure(ii) so that the air occupies one-third of the length of the cylinder it previously occupied.
Answers
Answer:
Let the piston be displaced by small amount X and the increase in pressure be △p.
Initial pressure P1=P0+AMg
Volume V1=V0
Final pressure P2=P1+△p
Volume V2=V0−△V,△V=Ax
Applying Poisson's equation
P1V1γ=P2V2γ
P1V1γ=(P1+△p)(V0−△V)γ
=P1V0γ(1+P1△P)(1−V0△V)γ
1=(1+P1△P)(1−γ
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Given,
The piston has been pushed as shown in figure(ii) so that the air occupies one-third of the length of the cylinder it previously occupied.
To find,
What would happen to;
(i) number of molecules
(ii) volume of the air
(iii) pressure of the air
(iv) What is the normal atmospheric pressure in terms of the height of mercury.
Solution,
We can solve the question by simply understanding and observing the figure given.
(i) As it is an air-tight piston, hence the number of molecules of air will remain constant.
(ii) The volume of the air will increase as there is more space now.
(iii) The pressure will decrease for the air molecules.
(iv) The normal atmospheric pressure in terms of the height of mercury will be 760 mm.
