Gases Solved Exercise
Solved exercise for chapter 3 chemistry book 1 for Fsc Pre engineering and Pre Medical and for entry test preparation for Engineering Universities and Medical Colleges.
EXERCISE
Q1: Select the correct answer out of the following alternative suggestions:
(i) Pressure remaining constant, at which temperature the volume of a gas will become twice of what it is at 0oC .
a. 546oC b.200oC c. 546 k d. 273k
(ii) Number of molecules in one dm3 of water is close to
a. b. c. d.
Hint: 1 dm3 of H2O =1000cm3 of H2O; 1000cm3 of H2O =1000g of H2O
No of moles of H2O =moles ( 1 ml of H2O =1 f of H2O)
(iii) Which of the following will have the same number of molecules at STP?
a. 280cm3 of CO2 and 280 cm3 of N2 O
b. 11.2dm3 of O2 and 32 g of O2
c. 44g of CO2 and 11.2 dm3 of CO
d. 28 g of N2 and 5.6 dm3 of oxygen
(iv) If absolute temperature of a gas is doubled and the pressure is reduced to one half , the volume of the gas will
a. remain unchanged b. Increase four times
c. reduce to d. be doubled
Hint: PV = RT, V=: V
(v) How should the conditions be changed to prevent the volume of a given gas from expanding when its mass is increased?
a. Temperature is lowered and pressure is increased.
b. Temperature is increased and pressure is lowers.
c. Temperature and pressure both are lowered
d. Temperature and pressure both are increased.
(vi) The molar volume of CO2 is maximum at
a. STP b. 127oC and 1 atm
c. 0oC and 2 atm d. 273 oC 1 atm
(vii) The order of the rate of diffusion of gases NH3, SO2 C12 and CO2 si :
a. NH3 >SO2 > C12 > CO2
b. NH3 > CO2 > SO2 > C12
c. C12 >SO2 > CO2 >NH3
(viii) Equal masses of methane and oxygen are mixed in an empty container at 25 oC . The fraction of total pressure exerted by oxygen is
a. b.
c. d.
Hint: In an empty container, the partial pressure of gas is directly proportional to the mole-fraction of the gas. Partial pressure mole fraction, at STP.
(viii) Gases deviate from ideal behavior at high pressure. Which of the following is correct for non-ideality?
a. At high pressure, the ga molecules move in one direction only.
b. At high pressure, the collision between the gas molecules are increased manifold.
c. At high pressure, the intermolecular attractions become significant.
d. At high pressure, the intermolecular attractions become significant.
(ix) The deviation of a gas from ideal behavior is maximum at
a. -10oC and 5.0atm b. -10oC and 2.0atm
c. 100 oC and 2.0 atm d. 0.oC and 2.0atm
(xi) A real gas obeying van der Waals equation will resemble ideal gas if
a. both ‘a’ and ‘b’ are larger b. both ‘a’ and ‘b’ are small
c. ‘a’ is small and ‘b’ is larger d. ‘a’ is larger and ‘b’ is small
Ans: (i)c (ii)d (iii)a (iv)b (v)a (iv)b (vii)b (viii)a (ix)d (x)a (xi)b
Q2. Fill in the blanks.
(i) The product of PV has the S.I. unit of _______.
(ii) Eight grams each of O2 and H2 at 27oC will have total K.E in the ratio of ________.
(iii) Smell of the cooking gas during leakage from a gas cylinder is due to of the property of _______of gases.
(iv) Equal _________of ideal gases at the same temperature and pressure contain ________number of molecules.
(v) The temperature above which a substance exists only as gas is called _____.
Ans. (i) atm dm3 (ii) :16 (iii) diffusion (iv) Volumes: equal (v) Critical temperature
Q3. label the following sentences as true or false.
(i) Kinetic energy of molecules is zero at 0oC.
(ii) A gas in a closed container will exert much higher pressure at the bottom due to gravity than at the top.
(iii) Real gases show ideal gas behavior at low pressure and high temperature.
(iv) Liquefaction of gases involves decrease in intermolecular spaces.
(v) An ideal gas on expansion will show Joule-Thomson effect.
Ans. (i) False (ii) False (iii) True (iv) False (v) False
Q4: (a) What is Boyle’s law of gases? Give its experimental verification.
(b) What are isotherms? What happen to the positions of isotherms when they plotted at high temperature for a particular gas.
(c) Why do we get a straight line when pressures exerted on a gas are plotted against inverse of volumes. This straight line changes its position in the graph by varying the temperature. Justify it.
(d) How will you explain that the value of the constant k in equation PV=k depends upon
(i) The temperature of the gas (ii) the quantity of the gas.
Ans: (a) Isotherms: The P-V curves obtained at constant temperature are called isotherms. These curves are obtained by plotting a graph between pressure on the x-axis and volume on the y-axis. Similar curves are obtained at fixed temperatures.
When the isotherms (P-V curves are plotted at high temperature, they go away from the axes. The reason is that, at higher temperature, the volume of the gas increase.
Ans.(c) A plot of P versus gives a straight line at constant temperature. This shows that p is directly proportional to . This straight line will meet at the origin where both P and are zero. The P goes down as the gas expands, falling away to zero as the volume approaches infinity (==0)
This straight line changes its position because both pressure and volume varies on varying the temperature. When temperature is increased both pressure and volume will increase. Keeping T constant and plotting P versus another straight line is obtained. This straight line goes away from x-axis. However , when temperature is decreased both the values of P and V will decrease. Again a straight line is obtained. This straight line will be closer to the x-axis
Ans.(d) General equation: PV=nRT _______(1)
The general gas equation contains the Boyle’s law, for which nRT are constant at fixed T and n.
Bolyle’s law: PV=k _______(2)
Form Eq (1) and Eq (2), we get
K=nRT
K=
K=constant x mT (at fixed R and M )
K mT _______(3)
This relation indicates that
(i) k T; it means k depends upon the temperature of the gas
(ii) k m; it means k depends upon the quantity of the gas.
Q5. (a) What is the Charles’s law? Which scale of temperature is used to verify that = k (pressure and number of moles are constant)
(b) A sample of carbon monoxide gas occupies 150.0mL at 25.0o C. it is then cooled at constant pressure until it occupies 100.0mL. What is the new temperature?
(c) Do you think that the volume of any quantity of a gas becomes zero at -273oC. Is it not against the law of conservation of mass? How do you deduce the idea of absolute zero from this information?
Ans.(a) The relation, =k can be verified only when T is taken on the Kelvin scale.
Ans.(b) V1 =150 mL T1 =273+25=298K
V2 =100 mL T2 =?
Formula Used:
=
or
T2 =
T2 =
T2 =198.67k
T2 =273+toC
toC=T2 – 273
toC=198.67-273
=-74.33oC Answer
Ans.(c) We know that : V1 =Vo (1+)
At -273oC, V-273 = Vo (1-)= Vo (1-1) =Vo x0=0
The volume of the gas become zero at- 273oC. But it is impossible to imagine that a
Gas which is matter occupies no space. It goes against the law of conservation of mass.
Hence, it follows that -273oC is the lowest temperature which a body can ever have. So -273o C is called the absolute zero of temperature. The scale of temperature on which -273oC is taken as zero is called absolute scale of temperature.
Q6. (a) What is Kelvin scale of temperature ? Plot a graph for one mole of an ideal gas to prove that a gas become liquid, carlier than -273o C .
(b) Throw some light on the factor in Charles’s law.
Ans.(b) In Charles’s law, the factor (0.00366 ) is the coefficient of expansion of given mass of gas at constant pressure. It shows that a gas expands by parts of its volume at 0o C for a rise of temperature of 1oC.
Statement of Charles’s Law: “At constant pressure, the volume of a given mass of gas increases or decreases by of its volume at0oC for every 1oC rise or fall in temperature .”
Mathematically.
Vt -=Vo +
It means that if we have 273 cm3 of gas at 0oC, its volume will increase by 1 cm3 for every 1oC rise in temperature if it is heated at constant pressure Thus,
At 1oC , the volume will become 274 cm3.
At 2oC , the volume will become 271 cm3.
At 273oC , the volume will become 0 cm3.
Similarly, if the gas is cooled by 1oC at constant pressure, its volume will decrease by 1 cm3 . Thus,
At -1oC , the volume of the gas becomes 272 cm3.
At -2oC , the volume of the gas becomes 275 cm3.
At -273oC , the volume of the gas becomes 546 cm3.
Q7. (a) What is the general gas equation? Derive it in various forms.
(b) Can we determine the molecular mass of an unknown gas if we know the pressure, temperature and volume along with the mass of that gas?
(c) How do you justify from general gas equation that increase in temperature or decrease of pressure decreases the density of gas?
(d) Why do we feel comfortable in expressing the densities of gases in the units of g dm-3rather than g cm-3, a which is used to express the densities of liquids and solids.
Ans.(b) Yes, we can determine the molecular mass of an unknown gas if we know its, P,T,V and m by applying the following formula:
M=
Ans.(c) We know form general gas equation,
d=
d=constant x (M and R being constant)
d=
Density is directly proportional to pressure and inversely proportional to temperature or decrease of pressure, decreases the density of the gas.
Ans.(d) The densities of gases are very low. They are about 1000 times smaller than the densities of liquids and solids. So, if gas densities are expressed in g cm-3, then the values will be very small. They may go to fourth place of decimal for some gases. When we express the densities in g dm-3, then the values of the densities become reasonable to be expressed. For example, the density of CH4 gas is 0.7168 g dm-3 at STP, but if it is expressed in g cm-3 , then it is 0.0007168 g cm-3 at STP. Therefore, we feel comfortable in expressing the densities of gases in the units of g dm-3 rather than in g cm-3.
Q8. Derive the units for gas constant R in general gas equation:
(a) When the pressure is in atmosphere and volume in dm3.
(b) when the pressure is in Nm-2 and volume in m3 .
( c) when energy Is expressed in ergs.
Q9. (a) what is Avogadro’s law of gases?
(b) Do you think that 1 mole of H2 and 1 mole NH3 at 0oC and 1 atm pressure will have equal Avogadro’s number of particles? If not, why?
(c) Justify that 1 cm3 of H2 and 1 cm3 of CH4 at STP will have the same number of molecules, when one molecules of CH4 is 8 times heavier than that of hydrogen.
Ans (b) 1 mole of H2 and 1 mole of NH3 at 0oC and 1 atm pressure will have equal number of molecules under the same conditions of temperature and pressure. Hence, 1 cm3 of H2 and 1 cm3 of CH4 at STP will have the same number of molecules.
Q10. (a) Dalton
(c) Derive an equation to find out the partial pressure of a gas knowing the individual moles of component gases and the total pressure of the mixture.
(d) Explain that the process of respiration obeys the Dalton
(e) How do you differentiate between diffusion and effusion? Explain Graham’s law of diffusion.
Ans. (a) For Dalton
Q11. (a) What is critical temperature of a gas? What is its importance for
Liquefaction of gases? Discuss Linde’s method of liquefaction of gases.
(b) What is Joule-Thomson Effect? Explain its importance in linde’s method of liquefaction of gases.
Ans. (a) Importance of Critical temperature for liquefaction of gases.
The critical temperature of the gases provides us the information about the
Condition under which gases liquefy. For example,O2 ,has a critical temperature 154.4k(-118,75 oC). It must be cooled below this temperature before it can be liquefied by applying high pressure.
Q12. (a) What is Kinetic molecular theory of gases? Give its postulates.
(b) How dose Kinetic molecular theory of gases explain the following gas laws:
(i) Boyle’s law (ii) Charles’s law
(iii) Avogadro’s law (iv) Graham’s law of diffusion.
Q13. (a) Gases show non-ideal behavior at low temperature and high pressure.
Explain this with the help of a graph.
(b) Do you think that some of the postulates of Kinetic molecular theory of gases are faulty? Point out these postulates.
(c) Hydrogen and helium are ideal at room temperature, but SO2 and C12are non ideal. How will you explain this?
Ans. (c) Hydrogen (b.p-253oC) and helium b.p-269oC)have a very low boiling points .They are far away from their boiling points at room temperature. Also, they have smaller number of electrons in their molecules and smaller molecular sizes, i.e., molecular weight. So, intermolecular forces are negligible at room temperature. Hence, they behave as an ideal gases at room temperature.
On the other hand, SO2 (b.p-10oC) and C12 (b.p-34oC) have boiling points near to room temperature. They are not far away from their boiling points at room temperature. Also, they have larger number of electrons in their molecules and larger molecular sizes. So, sufficient intermolecular attractive forces are present at room temperature. Hence, they behave as non-ideal at room temperature.
Q14. (a) Derive van der Waals equation for real gases.
(b) What is the physical significance of van der Waals constant, ‘a’ and ‘b’.
Give their units.
Ans. (b) Physical Significance of van der Waals constant ‘a’ and ‘b’
(i) Significance of’ a’: The value of constant ‘a’ is a measure of the intermolecular attractive forces and greater will be the ease of its liquefaction.
Units of ‘a’ :
The units of ‘a’ are related to the units of pressure, volume and number of moles.
P= or
a= =
a= atm dm6 mol-2
In SI units: a= ==Nm4 mole -2
(iii) Significance of ‘b’: The value of constant ‘b’ us related to the size of the molecule. Larger the size of the molecule, lager is the value of ‘b’. It is effective volume of the gas molecules.
Units of ‘b’:
‘b’ is the compressible volume per mole of gas. So the units of ‘b’ are related to the units of volume and moles.
V=nb or b =
b==dm3 mol-1
In SI units: b===dm3 mol-1
Q15. Explain the following facts:
(a) The plot of PV versus P is a straight line at constant temperature and with a fixed number of moles of an ideal gas.
(b) The straight line in (a) is parallel to pressure-axis and goes away from the pressure axis at higher pressure for many gases.
(c) The van der walls constant ‘b’ of a gas is four times the molar volume of that gas
(d) Pressure of NH3 gas at given conditions (say 1 atm pressure and room temperature) is less as calculated by van der Waals equation than that calculated by general gas equation.
(e) Water vapors do not behave ideally at 273 k.
(f) SO2 is comparatively non-ideal at 273 k but behaves ideally at 327 K.
Ans. (a) At constant temperature and with a fixed number of moles of an ideal gas, when the pressure of the gas is varied, its volume changes, but the product PV remains constant. Thus,
P1 V1 = P2 V2 = P3 V3 =
Hence, for any fixed temperature, the product PV when plotted against P.a straight line parallel to P-axis is obtained. This straight line indicates that PV remains constant quantity.
Ans. (b) Now, increase the temperature of the same from T1 to T2 .At constant temperature T2 and with the same fixed number of moles of an ideal gas, when the constant. However, the value of PV increase with increase in temperature. On plotting graph between P on x-axis is obtained. This straight line at T2 will be away from the x-axis. This straight line also shows that PV is a constant quantity.
Ans. (c) Excluded volume, ‘b’ is four times the molar volume fo gas. The excluding with each other as shown in Fig. The spheres are considered to be non-compressible. So the molecules cannot approach each other more closely than the distance, 2r . Therefore, the space indicated by the dotted sphere having radius, 2r will not be available to all other molecules of the gas. In other words, the dotted spherical space is excluded volume per pair of molecules.
Let each molecules be a sphere with radius =r
Volume of one molecules (volume of sphere)=
The distance of the closest approach of 2 molecules =2 r
The excluded volume for 2 molecules=
The excluded volume for 1 molecule=
=
=4Vm =b
The excluded volume for ‘n’ molecules=n b
Where Vm is the actual volume of a molecule.
Hence, the excluded volume or co-volume or non-compressible volume is equal to 4 times the actual volume of the molecules of the gas.
Ans. (d) The pressure of NH3 calculated by general gas equation is high because it is considered as an ideal gas. In an ideal gas, the molecules do not exert any force of attraction on one another. On the other hand, when the pressure of NH3is considered as a real gas. Actually, NH3 is a real gas. It consists of polar NH3 molecules approaches the walls of the container, it experiences an inward pull. Clearly, the molecule strikes the wall with a lesser force than it would have done it these are no attractive forces. As a result of this , the real gas pressure is less than the ideal pressure.
Ans. (e) Water vapors present at 273K do not behave ideally because polar water molecules exert force of attraction on one another.
Ans. (f) At low temperature, the molecules of SO2 possess low kinetic energy. They come close to each other. The e intermolecular attractive forces become very high. So, it behave non-ideally at 273K. At high temperature, the molecules of SO2 have high kinetic energy. The molecules are at larger distances from one other another. The intermolecular attractive forces become very weak. So, it behaves ideally at 327K.
Q16. Helium gas in a 100 cm3 container at a pressure of 500 torr is transferred to a container with a volume of 250cm3.What will be the new pressure
(a) if no change in temperature occurs
(b) if its temperature changes from 20 oC to 15 oC?
Solution:
Q17. (a) What are the densities in kg/m3 of the following gases at STP
(i) Methance, (ii) oxygen (iii) hydrogen
(P=101325Nm-2, T=273k, molecular masses are in kg mol-1 )
(b) Compare the values of densities in proportion to their mole masses.
(c) How do you justify that increase of volume upto 100 dm3 at 27 oC of 2 moles of NH3 will allow the gas behave ideally.
Solution:
Q18. A sample of krypton with a volume of 6.25 dm3 , a pressure of 765 torr and a temperature of 20oC is expanded to a volume of 9.55 dm3 and a pressure of 375 torr. What will be its final temperature in oC?
Solution:
Q19. Working at a vacuum line, a chemist isolated a gas in a weighing bulb with a volume of 255 cm3 , at a temperature of 25 oC and under a pressure in the bulb of 10.0torr. The gas weight 12.1 mg. what is the molecular mass of this gas?
Solution:
V=255 cm3 =0.255 dm3
Q20. What pressure is exerted by a mixture of 2.00g of H2 and 8.00g of N2 at 273 K in a dm3 vessel?
Solution:
Given: V=10dm3 ; T=273k
R=0.0821 dm3 atm K-1 mol-1
P H2 =? PN2 =?
Mass of H2 =2.00g Mass of N2 =8.00g
Molar mass of H2 =2g Molar mass of N2 =28g mol-1
nH2 ==1 mole ;n N2 ==0.286 mole
n=mH2 +nN2 =1+0.286=1.286 moles
PV=nRT
P x10 dm3 =1.286 molx 0.0821dm3 atm k-1 mol-1 x 273K
P=
P=2.88 atm Answer
Q21. (a) The relative densities of two gases A and B are 1:1.5 find out the volume of B which will diffused in the same time in which 150 dm3 of A will diffuse?
(b) Hydrogen (H2 ) diffuses through a porpous plate at a rate of 500cm3 per minute at 0oC. What is the rate of diffusion of oxygen through the same porpous plate 0oC?
(c) The rate of effusion of an unknown gas A through a pinhole is found to be 0.279 times the rate of effusion of H2 gas through the same pinhole.
Calculate the molecular mass of the unknown gas at STP.
Solution:
Q22. Calculate the number of molecules and the number of atoms in the given amounts of each gas
(a) 20cm3 of cH4 at 0oC and pressure of 700 mm of mercury
(b) 1 cm3 of NH4 at 100oC and pressure of 1.5 atm
Solution:
Q23. Calculate the masses of 1020 molecules of each of H2 ,O2 and CO2 at STP. What will happen to the masses of these gases, when the temperature of these gases are increased by 100oC and pressure is decreased by 100 torr.
Solution:
(a) Given: Molecules of H2 =1020
Now , 6.02x1023 molecules of H2 at STP =1mole =2g
1020 molecules of H2 at STP =
=0.332x10-5 g
=3.32x10-4 g Answer
(b) Given: Molecules of O2 =1020
Now, 6.02x1023 molecules of O2 at STP =32g
1020 molecules of CO2 STP =
=5.32x 10-3 g Answer
(c) Given: Molecules of CO2 =1020
Now, 6.02x1023 molecules of CO2 at STP =44g
1020 molecules of CO2 at STP =
=7.30x10-3 g Answer
Q24. (a) Two moles of NH3 are enclosed in a 5dm3 flask at 27oC. calculate the pressure exerted by the gas assuming that
(i) it behaves like an ideal gas
(ii) it behaves like a real gas
a=1.17 atm dm6 mol-2
b=0.0371 dm3 mol-1
(b) Also calculate the amount of pressure lessened due to forces of attractions at these conditions of volume and temperature.
(c) Do you expect the same decreased in the pressure of 2 moles of NH3having a volume of 40dm3 and at temperature of 27oC.
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