IISER Physics - Kinetic Theory of Gases
Exam Duration: 45 Mins Total Questions : 30
Real gas behaves as an ideal gas at
- (a)
high temperature and high pressure
- (b)
low temperature and high pressure
- (c)
low pressure and high temperature
- (d)
low pressure and low temperature
Real gas behaves as an ideal gas at low pressure and high temperature.
One mole of an ideal gas undergoes a process in which T=T0+qv3, where T0 and a are positive constants and v is volume.The volume for which pressure will be minimum
- (a)
\(({T_0\over2a})^{1/3}\)
- (b)
\(({T_0\over3a})^{1/3}\)
- (c)
\(({a\over2T_0})^{2/3}\)
- (d)
\(({a\over3T_0})^{2/3}\)
As, we know, ideal gas \(T={pV\over R}for\ n=1\)
\(\therefore\ \ {pV\over R}=T_0+aV^3\ or\ p={RT_0\over v}+aRV^2\)
For p to be minimum, \(dp\over dv\)=0
\(\therefore\ \ 0=-{RT_0\over V^2}+2aRV\)
\(\therefore\ \ v=({T_0\over2a})^{1/3}\)
Math the laws given in column I with their formulae given in column II and select the correct option from the choice given below
Column I | Column II |
A.Charle's law | 1.pV=constant[at constant temperature] |
B.Gay-Lussac's law | 2.\(V\over T\)=constant[At constant pressure] |
C.Boyle's law | 3.\(p\over T\)=constant[At constant volume] |
- (a)
A B C 1 2 3 - (b)
A B C 2 3 1 - (c)
A B C 3 2 1 - (d)
A B C 2 1 3
A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heats\(\gamma\).It is moving with speed v and is suddenly brought to rest.Assuming no heat is lost to the surroundings, its temperature increases by
- (a)
\({(\gamma-1)\over2R}Mv^2K\)
- (b)
\({(\gamma-1)\over2(\gamma+1)R}Mv^2K\)
- (c)
\({(\gamma-1)\over2\gamma R}Mv^2K\)
- (d)
\({\gamma Mv^2\over2R}K\)
Work done (W) in bringing the vessel at rest is equal to change in internal of gas (△U)
So, W=△U
\(\Rightarrow \frac{1}{2}mv^{2}=nC_v dT\)
\(\Rightarrow \frac{1}{2}mv^{2}=\frac{m}{M}\frac{R}{\gamma-1}dT\)
\(\Rightarrow dT=\frac{M(\gamma-1)v^{2}}{2R}\) kelvin
\(dT=\frac{\gamma-1}{2R}Mv^{2}\) kelvin
1 kg of diatomic gas is at a pressure of 8x144Nm-2.The density of the gas is 4kgm-3.What is the energy of the gas due to its thermal motion?
- (a)
3x104J
- (b)
5x104J
- (c)
6x104J
- (d)
7x104J
Thermal energy corresponds to internal energy
Mass=1kg
Density=4kgm-3
\(\Rightarrow\ \ Volume-{mass\over density}={1\over4}m^3\)
Pressure=\(8\times10^4Nm^{-2}\)
Internal energy=\({5\over2}p\times v=5\times 10^4J\)
If Cp and CV denote the specific heats of nitrogen per unit mass at constant pressure and constant volume respectively, then
- (a)
Cp-CV=\(R\over28\)
- (b)
Cp-CV=\(R\over14\)
- (c)
Cp-CV=R
- (d)
Cp-CV=2R
According to Mayer's relation,
Cp-CV=\(R\over m\)=\(R\over28\)
1 mole of a gas with \(\gamma={7\over5}\)is mi9xed with 1 mole of a gas with \(\gamma={5\over3},\)then the value of \(\gamma\) for the resulting mixture is
- (a)
\(7\over5\)
- (b)
\(2\over5\)
- (c)
\(24\over16\)
- (d)
\(12\over7\)
Using the relation,
\(\frac{n_1+n_2}{\gamma-1}=\frac{n_1}{\gamma_1-1}+\frac{n_2}{\gamma_2-1}\)
\(\Rightarrow \frac{1+1}{\gamma-1}=\frac{1}{(\frac{5}{3}-1)}+\frac{1}{(\frac{7}{5})-1}\)
or \(\frac{2}{\gamma-1}=\frac{3}{2}+\frac{5}{2}=4\)
\(\Rightarrow \gamma=\frac{3}{2}=\frac{24}{16}\)
Which of the following gases possesses maximum rms velocity, all being at the same temperature?
- (a)
Oxygen
- (b)
Air
- (c)
Carbon dioxide
- (d)
Hydrogen
For all gases all the same temperatures,
\({ v }_{ rms }\infty \frac { 1 }{ \sqrt { M } } \)
So, vrms is maximum for the lightest gas, i.e., hydrogen.
At room temperature the nns speed of the molecules of a certain diatomic gas is found to be 1930 m/s; the gas is:
- (a)
hydrogen
- (b)
fluorine
- (c)
oxygen
- (d)
chlorine
On any planet, the presence of atmosphere implies; (vrms = root mean square velocity of molecules and ve = escape velocity))
- (a)
vrms <<ve
- (b)
urms > ve
- (c)
vrms = ve
- (d)
vrms = 0
To expel half the mass of air from a large flask at 27° C it must be heated to:
- (a)
540C
- (b)
1770C
- (c)
2770C
- (d)
3270C
\(PV=\frac { m }{ M } \left( R\times 300 \right) ,PV=\frac { \left( m/2 \right) }{ M } RT\)
Hence, T' = 600 K or t = 3270 C
We have ajar A filled with gas characterised by parameters P, V and T and another jar B filled with gas with parameters 2P, V/4 and 2T, when the symbols have their usual meanings. The ratio of the number of molecules of jar A to those of jar B is:
- (a)
1:1
- (b)
1:2
- (c)
2:1
- (d)
4:1
PV=\(\frac { m }{ M } \)RT
where m is the mass of gas and M, the molecular mass
If mo is the mass of one molecule and N number of molecules, then m = m0 N
and PV=\(\frac { { m }_{ 0 }N }{ M } \)RT
For jar A: PV=\(\frac { { m }_{ 0 }{ N }_{ 1 } }{ M } \)RT
For jar B: (2P) \(\left( \frac { V }{ 4 } \right) =\frac { { m }_{ 0 }{ N }_{ 2 }R(2T) }{ M } \)
∴ \(\frac { { N }_{ 1 } }{ { N }_{ 2 } } =\frac { 4 }{ 1 } \)
A container has N molecules at absolute temperature T. If the number of molecules is doubled but kinetic energy in the box remains the same as before, the absolute temperature of the gas is:
- (a)
T
- (b)
T/2
- (c)
2T
- (d)
zero
\(\bar { KE } =\frac { 3 }{ 2 } \)kT
On doubling the number of molecules and keeping total KE same, average KE (\(\bar { KE } \)) becomes half, resulting in half the temperature.
The mass of oxygen gas occupying a volume of 11.2 litres at a temperature 27°C and a pressure of 760 mm of mercury is: (molecular weight of oxygen = 32)
- (a)
0.001456 kg
- (b)
0.01456 kg
- (c)
0.1456 kg
- (d)
1.1456 kg
At S.T.P:
P1 = 760 mm; V1 = 22.4 litres, m1 = 32 g and T1 = 273 K
Given, P2 = 760 m; V2 = 11.2 liters; T2 = 27oC = 300 K
\(\therefore \frac { 760\times 22.4 }{ 32\times 273 } =\frac { 760\times 11.2 }{ { m }_{ 2 }\times 300 } \)
\(\therefore { m }_{ 2 }=14.56g=0.01456kg.\)
Half a mole of helium at 27° C and at a pressure of 2 atmosphere is mixed with l.5 mole of Nj at 77° C and at a pressure at 5 atmosphere so that the volume of the mixture is equal to the sum of their initial volumes. If the temperature of the mixture is 69° C, its pressure is:
- (a)
3.5 atm
- (b)
3.8 atm
- (c)
3.95 atm
- (d)
4.25 atm
P1V1 = u1RT1
or V1 = \(\frac { { u }_{ 1 }{ RT }_{ 1 } }{ { P }_{ 1 } } =\frac { 1 }{ 2 } \frac { R\left( 300 \right) }{ 2 } =75R\)
P2V2 = u2RT2
or V2 = u2\(\frac { R{ T }_{ 2 } }{ { P }_{ 2 } } =1.5\frac { R(350) }{ 5 } =105R\)
P(V1 + V2) = (u1 + u2)Rt
or P(75R + 105R) = \(\left( \frac { 1 }{ 2 } +1.5 \right) \) R (273 + 69)
or P x 180R = 2 x R x 342
P = \(\frac { 342 }{ 90 } \) = 3.8 atm
During an experiment an ideal gas is found to obey an additional law VP2 = constant. The gas is initially at temperature T and volume V, when it expands to volume 2V, the resulting temperature is
- (a)
T/2
- (b)
2T
- (c)
\(\sqrt{2}\)T
- (d)
T/\(\sqrt{2}\)
PV=RT=constant ....(i)
Also, VP2=constant ....(ii)
From eqn (i), P=\(\frac { RT }{ V } \)
From eqn (ii), V\(\frac { RT }{ V } \)=constant
\(\frac { RT }{ V } \)=constant or \(\frac { { T }^{ 2 } }{ { V } } \)=constant
∴ \(\frac { { T }^{ 2 } }{ { V } } =\frac { { T }^{ '2 } }{ { V } } \) or T'2=2T2
∴ T'= \(\sqrt{2}\)T
At which of the following temperatures would the molecules of a gas have twice the average kinetic energy they have at 27° C?
- (a)
313°C
- (b)
373°C
- (c)
393°C
- (d)
586°C
Which of the following methods will enable the volume of an ideal gas to be made four times greater? (Consider absolute temperature)
- (a)
Quarter the pressure at constant temperature
- (b)
Quarter the temperature at constant pressure
- (c)
Half the temperature, double the pressure
- (d)
Double the temperature, double the pressure
Under which of the following conditions is the law PV = RT obeyed most closely by a real gas?
- (a)
High pressure and high temperature
- (b)
Low pressure and low temperature
- (c)
Low pressure and low temperature
- (d)
High pressure and low temperature
At low pressure and high temperature, the molecules are farther apart so that molecular size is negligible as compared to the size of the vessel and also molecular forces do not come in
Consider I cc sample of air at absolute temperature To at sea level and another 1 cc sample of air at a height, where pressure is one-third atmosphere.The absolute temperature T of the sample at the height is:
- (a)
equal to (T0 / 3)
- (b)
equal to (3 / T0)
- (c)
equal to To
- (d)
cannot be determind in terms of T0 from the above data
Mass of 1 cc of gas at sea level and at given height does not remain the same and gas laws hold good only for a constant mass.
The air density at Mount Everest is less than that at sea level. It is found by mountaineers that for one trip lasting a few hours, the extra oxygen needed by them corresponds to 30,000 cc at sea level (pressure = 1 atmosphere, temperature = 27°C). Assuming that the temperature around Mount Everest is -73° e and that the oxygen cylinder has capacity of 5.2 litres, the pressure at which oxygen be filled (at site) in the cylinder is:
- (a)
3.86 atm
- (b)
5.00 atm
- (c)
5.77 atm
- (d)
I atm
The root mean square velocity of the molecules of a gas is 1260 m/s. The average speed of the molecules is
- (a)
1029 ms-1
- (b)
1161 ms-1
- (c)
1671 ms-1
- (d)
917 ms-1
vav=\(\sqrt { \frac { 8kY }{ m\pi } } \), vrms=\(\sqrt { \frac { 3kT }{ m } } \)
∴ \(\frac { v_{ av } }{ { v }_{ rmp } } =\frac { \sqrt { 8/\pi } }{ \sqrt { 3 } } \)
or \(\frac { v_{ av } }{ 1260 } =\sqrt { \frac { 8 }{ 3\pi } } \)
∴ vav=\(1260\times \sqrt { \frac { 8 }{ 3\pi } } \)=1161 ms-1.
The quantity PV/kT(k= Boltzmann's constant) represents:
- (a)
number of moles of the gas
- (b)
total mass of the gas
- (c)
number of molecules in the gas
- (d)
density of the gas
\(\frac { PV }{ kT } =\frac { nRT }{ (R/N)T } \)=nN = number of molecules In the gas.
Two gases of equal mass are in thermal equilibrium. If Pa ,Pb and Va and Vb are their respective pressures and volumes, then which relation is true?
- (a)
PaVa = PbVb
- (b)
Pa/Va = Pb/Vb
- (c)
Pa = Pb; Va ≠ Vb
- (d)
Pa ≠ Pb; Va ≠ Vb
The equation of state corresponding to 8 g of O2 is
- (a)
PV = 8RT
- (b)
PV = \(\frac { RT }{ 4 } \)
- (c)
PV = RT
- (d)
PV = \(\frac { RT }{ 2 } \)
8 g of oxygen is equivalent to (1/4) mole
PV = uRT = \(\frac { RT }{ 4 } \)
Consider a gas with density \(p\) and \(\bar { c } \) as the root mean square velocity of its molecules contained in a volume. If the system moves as a whole with velocity u, then the pressure exerted by the gas is:
- (a)
\(\frac { 1 }{ 3 } p\bar { c } ^{ 2 }\)
- (b)
\(\frac { 1 }{ 3 } \rho \left( \bar { c } +v \right) ^{ 2 }\)
- (c)
\(\frac { 1 }{ 3 } \rho \left( \bar { c } -v \right) ^{ 2 }\)
- (d)
\(\frac { 1 }{ 3 } \rho \left( \bar { c } ^{ 2 }-v \right) ^{ 2 }\)
When temperature of an ideal gas is increased from 27°C to 227°C, its rms speed is changed from 400 m/s to vs. The Vs is:
- (a)
516 m/s
- (b)
450 m/s
- (c)
310 m/s
- (d)
746 m/s
The value of \(\frac { PV }{ T } \) for one mole of an ideal gas is nearly equal
- (a)
2.7 mol-1 k-1
- (b)
8.3 mol-1 k-1
- (c)
4.2 J mol-1 k-1
- (d)
2 cal mol-1 k-1
One litre of oxygen at a pressure of 1 atm and two litres of nitrogen at a pressure of 0.5 atrn are introduced into a vessel ofvolurne I L. If there is no change in temperature, the final pressure of the mixture of gas (in atm) is:
- (a)
1.5
- (b)
2.5
- (c)
2
- (d)
4
Ideal gas equation is given by
PV = nRT
For oxygen : P = 1 atm; V = 1L; n = \({ n }_{ { O }_{ 2 } }\)
Therefore, eqn (i) becomes.
\(\therefore\) 1 x 1 = \({ n }_{ { O }_{ 2 } }\) RT or \({ n }_{ { O }_{ 2 } }\) = \(\frac { 1 }{ RT } \)
For nitrogen p = 0.5 atm; V= 2L; n = mN
\(\therefore\) 0.5 x 2 = \({ n }_{ { N }_{ 2 } }\) RT or \({ n }_{ { N }_{ 2 } }\)= \(\frac { 1 }{ RT } \)
For mixture of gas:
PmixVmix = nmix RT
Here, nmix = no + \({ n }_{ { N }_{ 2 } }\)
\(\therefore\) \(\frac { { P }_{ mix }{ V }_{ mix } }{ RT } =\frac { 1 }{ RT } +\frac { 1 }{ RT } \)
or PmixVmix = 2 \(\therefore\)Vmix = 1L; Pmix = 2atm
The ratio of the vapour densities of two gases at a given temperature is 9 : 8 .The ratio of the rms velocities of their molecules is :
- (a)
3 : 2\(\sqrt { 2 } \)
- (b)
2\(\sqrt { 2 } \) : 3
- (c)
9 : 8
- (d)
8 : 9