11th Standard Syllabus & Materials
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Published on: 13/05/2022
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Questions + Answers key
Take MCQ Physics Test1.
Define and derive Boltzmann constant.
2.
State and derive the perfect or ideal gas equation?
3.
State and explain Charle's law.
4.
State and explain Boyle's law.
5.
Derive an expansions for the work done in one cycle during adiabetre expansion.
1.
It is the gas constant per molecule
\(\therefore { k }_{ B }=\frac { R }{ { N }_{ A } } or\ R={ k }_{ B }{ N }_{ A }\)
R - gas const ; NA - Avogadro No.;
KB - Boltzmann
No.of moIecu Ies n =\(\frac{\text {No.of molecules}}{\text {Avogadro's no}}\)=\(\frac{N}{N_A}\)
pv = nRT=\(\frac{N}{N_A}\)KBNAT
pv = KBNT
2.
This-equation gives the relation between pressure P, volume. (v) and absolute temperature (T) of a gas.
The equation is PV = nRT
n - number of molecules of the gas
R - universal gas constant .
Derivation:
Accumulate the Boyle's law, for a gn mass of a gas at constant temperature.
v\(\times\)\(\frac{1}{p}\) .............(1)
Accumulate to charle's law, for a gn mass of a gas at constant pressure,
V\(\times\)T ................(2)
Combining (1) & (2)
v\(\times\) \(\frac{1}{p}\) (or) v = constant \(\frac{T}{P}\)(or) \(\frac{pv}{T}\) = constant
constant is called universal gas constant R.
pv = RT.
For one molecule of a gas, the constant has same value for all gases.
For n moles of a gas pv = nRT.
This is perfect (or) ideal gas equation.
3.
It states that if the pressure remains const, then the v0 of a g mass of a gas increases or decreases by \(\frac{1}{2723.15}\) of its volume at 0oC for each 1°C rise or fall of temperature Vo - volume of the gn mass
of a gas at O°C
Accumulate to charle's law, its volume at 1°C is
\({ v }_{ 1 }={ v }_{ 0 }+\frac { { v }_{ 0 } }{ 273.15 } ={ v }_{ 0 }\left( 1+\frac { 1 }{ 273.15 } \right) \)
Volume of the gas at toC
\({ v }_{ 1 }={ v }_{ 0 }\left( 1+\frac { 1 }{ 273.15 } \right) \)
If To and T are temperature on kelvin scale corresponding to 0°C & toC, then
To= 27:3.15+0 = 273.15
T = 273.15 +t
\(\therefore { v }_{ t }={ v }_{ 0 }\left( \frac { T }{ { T }_{ 0 } } \right) or\frac { { v }_{ 1 } }{ T } =\frac { { v }_{ 0 } }{ { T }_{ 0 } } \)
\(\frac{v}{T}\) = constant i.e. v x T.
So, the law states that at constant pressure the volume of a gn mass of a gas is directly proportional to its absolute temperature. The graph. between. V and T for a gn mass of a gas at constant pressure is a St line.

4.
It states that the volume of a gn mass of a gas is inversely proportional to it pressure provided the temperature remains constant.
\(v\times \frac { 1 }{ P } (or)v=\frac { k }{ p } \) (or) pv = constant
Its value depends on
(i) mass of the gas
(ii) its temperature and
(iii) the units in which P and v are measured.
P1 & V1 - initial values of pressure and volume
P2 & v2 - Final values of pressure and volume
then accumulate the Boyle's law P1 V1 = P2 v2
Graph between P vs. V and P vs.\(\frac{1}{v}\) for a gn mass a gas a constant temperature T are shown below.

5.
Consider one mole of an ideal gas enclosed in a cylinder with perfectly non conducting piston.
P1 - initial pres.
V1 - initial volume.
T1 - Initial temperature
A - area of cross-section.
Force excited by the gas on the piston is F = Px A where P - pressure of the gas during expansion.
11th Standard Syllabus & Materials
11th Standard
TN 11th Tamil பீடு பெற நில் - செய்யுள் - காவடிச்சிந்து Important Questions And Answers Study Material - QB365 Set A
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