Study of Gas Laws Exe-7 Structured Answer Chemistry Class-9 ICSE Selina Publishers

Study of Gas Laws Exe-7 Structured Answer Chemistry Class-9 ICSE Selina Publishers Solutions Chapter-7. Step By Step ICSE Selina Concise Solutions of Chapter-7 Study of Gas Laws with All Exercise including MCQs, Very Short Answer Type, Short Answer Type, Long Answer Type, Numerical and Structured/Application Questions Solved . Visit official Website CISCE for detail information about ICSE Board Class-9.

Study of Gas Laws Exe-7 Structured Answer Chemistry Class-9 ICSE Concise Selina Publishers

Board ICSE
Publications Selina Publication
Subject Chemistry
Class 9th
Chapter-7 Study of Gas Laws
Book Name Concise
Topics Solution of Exercise – 7 Structured/Application Answer Type
Academic Session 2023-2024

E. Exercise – 7 Structured/Application Answer Type

Study of Gas Laws Class-9 Chemistry Concise Solutions  

Page-133

Question 1.

(a) State the law verified by the following figure:

(a) State the law verified by the following figure:

(b) Draw P.V. isothermal for the above law.

Answer:

(a) Boyle’s law — It states that the volume of a given mass of a dry gas is inversely proportional to its pressure at a constant temperature i.e., P1V1 = P2V2

(b) P.V. isothermal for Boyle’s law is shown below :

(b) P.V. isothermal for Boyle's law is shown below :

Question 2.

(a) State the law which the following graph verifies.
(a) State the law which the following graph verifies.

(b) Derive the mathematical expression for it.

(c) Give one application/use where the above law is employed.

Answer:

(a) Charles’ law — It states that pressure remaining constant, the volume of a given mass of dry gas increases or decreases by 1/273 of its volume at 0°C for each 1°C increase or decrease in temperature, respectively.

(b) Let V0 be the volume of a fixed mass of a gas at 0°C and let V be its volume at temperature t°C at constant pressure. Then according to Charles’ law,

V = Vo + (Vo/273)t (when P is constant)

V = Vo (1 + t/273)

= Vo (273 + t)/273

V =V0/273 T where T = 273 + t

For a given mass of a gas,

V0/273 = constant

∴ V = k x T (where k is constant)

or V α T and V/T = K

Suppose a gas occupies V1 cm3 at T1 temperature and V2 cm3 at T2 temperature, then by Charles law,

V1 α T1

or V1 = kT1 (k is constant)

or
V1/T1 = k and

V2 α T2

or V2/T2 = k

∴ V1/T1 = V2/T2 = k (at constant pressure)

(c) Volume of a given mass of a gas is directly proportional to its temperature, hence, density decreases with an increase in temperature. This is the reason that hot air is filled into balloons used for meteorological purposes.

Question 3.

Plot V versus absolute T (K) by the given data:

Temperature Volume in litres
27°C 4.8
77°C 5.6
127°C 6.4
177°C 7.2
227°C 8.0

(a) The graph between V and T is a …………… .

(b) Check whether the line passes through the origin.

(c) Which law is obeyed.

Answer:

Converting temperature to Kelvin Scale :

Temp
(°C)
Temp
(K)
Volume
(litres)
27 300 4.8
77 350 5.6
127 400 6.4
177 450 7.2
227 500 8.0

Graph of V versus absolute T is shown below :

Plot V versus absolute T (K) by the given data:

(a) The graph between V and T is a straight line.

(b) Yes, as we can see in the graph, the line passes through the origin.

(c) Charles’ law

—  : End of Study of Gas Laws Exe-7 Structured Answer Class-9 ICSE Chemistry Solutions :–

Return to  Return to Concise Selina ICSE Chemistry Class-9 

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