261.
Which one of the following is the wrong assumption of kinetic theory of gases ? A
Momentum and energy always remain conserved.
B
Pressure is the result of elastic collision of molecules with the container’s wall.
C
Molecules are separated by great distances compared to their sizes.
D
All the molecules move in straight line between collision and with same velocity.
Answer :
All the molecules move in straight line between collision and with same velocity.
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Molecules move very fast in all directions in a straight line by colliding with each other but with different velocity.
262.
Compressibility factor of a gas is given by the equation $$Z = \frac{{PV}}{{nRT}}.$$ On this basis, mark the correct statement. A
When $$Z > 1,$$ real gases get compressed easily.
B
When $$Z = 1$$ real gases get compressed easily.
C
When $$Z > 1,$$ realgasesare difficult to compress.
D
When $$Z = 1,$$ realgasesare difficult to compress.
Answer :
When $$Z > 1,$$ realgasesare difficult to compress.
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Discuss Question
When $$Z > 1,$$ real gases are difficult to compress because the forces of attraction between the molecules are very feeble.
263.
$$XmL$$ of $${H_2}$$ gas effuses through a hole in a container in 5 seconds. The time taken for the effusion of the same volume of the gas specified below under identical conditions is : A
$${\text{10 seconds : }}He$$
B
$${\text{20 seconds :}}\,{O_2}$$
C
$${\text{25 seconds :}}\,CO$$
D
$${\text{55 seconds :}}C{O_2}$$
Answer :
$${\text{20 seconds :}}\,{O_2}$$
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$$\eqalign{
& {\text{Under}}\,{\text{indential}}\,{\text{conditions,}}\,\frac{{{r_1}}}{{{r_2}}} = \sqrt {\frac{{{M_2}}}{{{M_1}}}} \cr
& {\text{As rate of diffusion is also inversely proportional to}} \cr
& {\text{time,}}\,{\text{we}}\,{\text{will}}\,{\text{have}},\,\frac{{{t_2}}}{{{t_1}}} = \sqrt {\frac{{{M_2}}}{{{M_1}}}} \cr
& \left( {\text{A}} \right)\,{\text{Thus,}}\,{\text{For}}\,He,\,\,{t_2} = \sqrt {\frac{4}{2}} \left( {5s} \right) = 5\sqrt {2s} \ne 10s; \cr
& \left( {\text{B}} \right)\,{\text{For}}\,{O_2},\,\,{t_2} = \sqrt {\frac{{32}}{2}} \left( {5s} \right) = 20s \cr
& \left( {\text{C}} \right)\,{\text{For}}\,CO,\,\,{t_2} = \sqrt {\frac{{28}}{2}} \left( {5s} \right) \ne 25s\,; \cr
& \left( {\text{D}} \right)\,{\text{For}}\,C{O_2},\,\,{t_2} = \sqrt {\frac{{44}}{2}} \left( {5s} \right) \ne 55s \cr} $$
264.
Two vessels of volumes $$16.4\,L$$ and $$5\,L$$ contain two ideal gases of molecular existence at the respective temperature of $${27^ \circ }C$$ and $${227^ \circ }C$$ and exert 1.5 and 4.1 atmospheres respectively. The ratio of the number of molecules of the former to that of the later is A
$$2$$
B
$$1$$
C
$$\frac{1}{2}$$
D
$$\frac{1}{3}$$
Answer :
$$2$$
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$$\eqalign{
& {\text{Given conditions}} \cr
& {V_1} = 16.4\,L,\,{V_2} = 5\,L \cr
& {P_1} = 1.5\,atm,\,{P_2} = 4.1\,atm \cr
& {T_1} = 273 + 27 = 300\,K, \cr
& {T_2} = 273 + 227 = 500\,K \cr
& {\text{Applying gas equation,}}\,\frac{{{P_1}{V_1}}}{{{P_2}{V_2}}} = \frac{{{n_1}{T_1}}}{{{n_2}{T_2}}} \cr
& \frac{{{n_1}}}{{{n_2}}} = \frac{{{P_1}{V_1}{T_1}}}{{{P_2}{V_2}{T_2}}} \cr
& \therefore \,\,\frac{{1.5 \times 16.4 \times 500}}{{4.1 \times 5 \times 300}} = \frac{2}{1} \cr} $$
265.
Three different gases $$X, Y$$ and $$Z$$ of molecular masses $$2, 16$$ and $$64$$ were enclosed in a vessel at constant temperature till equilibrium is reached.
Which of the following statement is correct? A
Gas $$Z$$ will be at the top of the vessel
B
Gas $$Y$$ will be at the top of the vessel
C
Gas $$Z$$ will be at the bottom and $$X$$ will be at the top
D
Gases will form homogenous mixture
Answer :
Gases will form homogenous mixture
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All the gases occupy the available volume and will form homogeneous mixture.
266.
The compressibility factor for an ideal gas is A
1.5
B
1.0
C
2.0
D
$$\infty $$
Answer :
1.0
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The compressibility factor of a gas is defined as $$Z = \frac{{p{V_m}}}{{RT}}$$
267.
A flask of capacity $$2\,L$$ is heated from $${35^ \circ }C$$ to $${45^ \circ }C.$$ What volume of air will escape from the flask ? A
10$$\,mL$$
B
20$$\,mL$$
C
60$$\,mL$$
D
50$$\,mL$$
Answer :
60$$\,mL$$
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$$\frac{{{V_1}}}{{{T_1}}} = \frac{{{V_2}}}{{{T_2}}};{V_1} = 2\,L,$$ $${T_1} = 35 + 273 = 308\,K$$
268.
What will be the volume of $$2.8\,g$$ of carbon monoxide at $$27{\,^ \circ }C$$ and 0.821 atmospheric pressure ? A
2.5$$\,L$$
B
4$$\,L$$
C
3.5$$\,L$$
D
3$$\,L$$
Answer :
3$$\,L$$
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$$\eqalign{
& n = \frac{m}{M} \cr
& \,\,\,\,\, = \frac{{2.8}}{{28}} \cr
& \,\,\,\,\, = 0.1 \cr
& V = \frac{{nRT}}{P} \cr
& \,\,\,\,\,\, = \frac{{0.1 \times 0.0821 \times 300}}{{0.821}} \cr
& \,\,\,\,\,\, = 3\,L \cr} $$
269.
Molecular mass of a gas is $$78\,\,g\,\,mo{l^{ - 1}}.$$ Its density at $$98{\,^ \circ }C$$ and $$1\,atm$$ will be A
$$200\,\,g\,\,{L^{ - 1}}$$
B
$$2.56\,\,g\,\,{L^{ - 1}}$$
C
$$256\,\,g\,\,{L^{ - 1}}$$
D
$$78\,\,g\,\,{L^{ - 1}}$$
Answer :
$$2.56\,\,g\,\,{L^{ - 1}}$$
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$$P = 1\,atm,T = 98 + 273 = 371\,K,$$ $$M = 78\,\,g\,\,mo{l^{ - 1}}$$
270.
By what factor does the average velocity of a gaseous molecule increase when the temperature (in kelvin) is doubled? A
2.8
B
4.0
C
1.4
D
2.0
Answer :
1.4
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$$\eqalign{
& {\text{Average velocity}} = \sqrt {\frac{{8RT}}{{\pi M}}} \cr
& \therefore \,\,{V_{av}} \propto \sqrt T \,\,or\,\,\frac{{{{\left( {{V_{av}}} \right)}_2}}}{{{{\left( {{V_{av}}} \right)}_1}}} \cr
& = \sqrt {\frac{{2T}}{T}} \cr
& = 1.4 \cr} $$