Book Store

Download books and chapters from book store.
Currently only available for.
CBSE Gujarat Board Haryana Board

Previous Year Papers

Download the PDF Question Papers Free for off line practice and view the Solutions online.
Currently only available for.
Class 10 Class 12

Derive an expression for the pressure of a gas.

The gas exerts pressure on the walls of container because molecules of gas are in random motion and strike with the walls of container and hence exert the thrust (force) on the wall. The thrust exerted by molecules per unit area of the wall of the container is called pressure.

Since molecule collide, with face L due to X-component of motion, therefore, due to collision only X-component of momentum will change. Since collision is elastic, therefore momentum after collision is,

stack straight p subscript 1 apostrophe with rightwards arrow on top space equals space minus mv subscript 1 straight x end subscript straight i with hat on top space plus space mv subscript 1 straight y end subscript straight j with hat on top plus mv subscript 1 straight z end subscript straight k with hat on top

Due to collision, the change in momentum of molecule is,

space stack straight p subscript 1 apostrophe with rightwards arrow on top minus stack straight p subscript 1 with rightwards arrow on top space equals space minus 2 mv subscript 1 straight x end subscript straight i with hat on top

Let n be the number of molecules per unit volume. Since the molecules approach towards face L with velocity v_1x, therefore, the molecules lying in the cylinder of length v_1x and area of cross-section A may collide with face L on are a in one second.
Since molecular motion is random, therefore, it is logical to suppose only half of the molecules that lie in cylinder will strike the area A in one second.
The gas exerts pressure on the walls of container because molecules o

According to Newton’s third law of motion ‘action and reaction are equal and opposite.’ Therefore gas molecules exert equal and opposite force on area A of face L as experienced by gas molecules.
Force on area A of face L is,

straight F subscript straight x equals nmA straight v with rightwards arrow on top subscript straight x squared

Therefore pressure on face L is,

<pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>

3837 Views

If the absolute temperature of gas is increased by 300%, then by what percentage the r.m.s. velocity of molecules will increase?

According to the kinetic interpretaion of temperature, r.m.s velocity is directly proportional to square root of absolute temperature.

i.e.,

v_{r} \propto \sqrt{T}

... (1)

Let the initial temperature of gas be T.

If the temperature is increased by 300% then the temperature of gas becomes 4T.

So, using equation (1), we have

v_{r}' \propto \sqrt{4 T}

... (2)

Therefore, from equations (1) and (2), we have

v_r' = 2v_rThe r.m.s. velocity at 4T temperature is double the r.m.s. velocity at temperature T.

So, percentage increase in velocity is given by,

% age = \frac{2 v - v}{v} \times 100 = 100 %

111 Views

Calculate the number of molecules in 1cc of ideal gas at 27°C and 10mm pressure of mercury. Mean kinetic energy of molecule at 27°C is 6.21x10^–14 ergs.

Given, 1 cc of ideal gas at 27^o C and 10 mm pressure of mercury.

Mean K.E of molecule = 6.21x10^–14 ergs.

We know that,

Pressure of a gas is equal to two-third of the average kinetic energy per unit volume of the gas.

Thats is,

P = \frac{2}{3} E_{d} \Rightarrow E_{d} = \frac{3}{2} P

Here we have,

P = 10 mm of mercury = 1 cm of mercury = 1 \times 13.6 \times 980 = 1.33 \times 10^{4} dyne / {cm}^{2} .

Therefore,

E_{d} = \frac{3}{2} P = \frac{3}{2} \times 1.33 \times 10^{4} = 2 \times 10^{4} erg / cc . . . . (1)

Let n be the number of molecules in 1 cc of gas, therefore the energy per cm³ of gas is,

E_{d} = n \times 6.21 \times 10^{- 14} ergs / cc

...(2)

From equations (1) and (2),

Number of molecules is =

n = \frac{2 \times 10^{4}}{6.21 \times 10^{- 14}} = 3.22 \times 10^{17} molecules

151 Views

A vessel of volume V = 5 litre contains 1.4gm of N₂ at temperature 1800K. If at this high temperature, 30% of the gas dissociates into atoms, then find the pressure of gas.

Given,

Volume of the vessel, V = 5

l

Molecular weight of nitrogen = 28 g

Temperature, T = 1800 K

Number of moles in 1.4 gm of N₂ is,

n_{o} = \frac{1.4}{28} = \frac{1}{20} moles

Also, given that at high temperature 30% of the gas dissociates into atoms.

Number of moles of gas dissociated into atoms is,

n' = 0.3 n_{o} = 0.3 \times \frac{1}{20} = \frac{3}{200} moles

Number of moles of atomic nitrogen is,

n_{1} = 2 n' = \frac{3}{200} \times 2 = 0.03 moles

Number of moles of molecular nitrogen is,

n_{2} = 0.7 \times \frac{1}{20} = 0.035 moles

Thus total number of moles of nitrogen gas is given by n = n₁+ n₂ = 0.065 moles

Now, using the ideal gas equation,
PV = nRT

We have, n = 0.065
T = 1800 K
V = 5 litre = 5

\times 10^{- 3}

m³

Therefore,

P r e s s u r e, P = \frac{nRT}{V} = \frac{0.065 \times 8.314 \times 1800}{5 \times 10^{- 3}} = 1.945 \times 10^{5} N / m^{2}

163 Views

Two containers of same size are at the same temperature. One of them contains 1 kg of H₂ gas and other 1 kg of O₂ gas.
(a) Which vessel contains more molecules?
(b) Which vessel is under greater pressure?

(a) Since molecular weight of H₂ is less than that of oxygen, therefore the number of moles and hence molecules in 1 kg of H₂ are greater than in O₂.

b) Pressure of a gas in terms of density is given by,

P =

\frac{1}{3} ρ v^{2}

Here, density of both gases is the same.

So,

P \propto v^{2}

Now, according to kinetic interpretation of temperature,

\frac{1}{2} m v^{2} = \frac{3}{2} K T

i.e., P

\propto v^{2} \propto \frac{1}{m}

Mass of hydrogen molecules is less than that of oxygen molecule. Therefore, more pressure is exerted on the H₂ gas container than the O₂ gas container.

138 Views

Chapter Chosen

Book Chosen

Subject Chosen

Book Store

Previous Year Papers

Kinetic Theory

Derive an expression for the pressure of a gas.