Q: यदि Z संपीड़न गुणक हो तो कम दाब पर वांडरवाल्स समीकरण को लिखा जा सकता है:

संपीडन गुणांक (Z)=RTPVएक मोल वास्तविक गैस के लिए वांडर वाल समीकरण(P+V2a)(V−b)=RTनिम्न दाब पर, आयतन बहुत अधिक होगा। अतः अत्यधिक आयतन V की तुलना में b को नगण्य मानने पर, अतःV−b≈V(P+V2a)V=RTPV+Va=RTPV=RT−VaRTPV=1−VRTaअत: Z=1−VRTa

Q: Which one, among the following, is the Van der Waals' equation, describing the behaviour of one mole of a real gas over wide ranges of temperature and pressure?

The van der Waals' equation for n moles of a gas is[p+V2n2a](V−nb)=nRT For one mole (n=1)(p+V2a)(V−b)=RT

Q: Volume occupied by molecules of one mole gas at NTP, each having radius of 10−8 cm is:

Volume of molecules in one mole =4×N×V=4×N×34πr3=4×6.023×1023×34×722×(10−8)3=10.09 mL

Q: At relatively high pressure, van der Waals' equation reduces to:

Van der waal equation is, (P+V2an2)(V−nb)=nRT…. for n moles of gas For 1 mole, the equation becomes, (P+V2a)(V−b)=RT At high pressure, the Pcorrection = negligible but volume correction cannot be regarded as negligible. P(V−b)=RT or PV−Pb=RT or PV=RT+Pb

Question 1

यदि Z संपीड़न गुणक हो तो कम दाब पर वांडरवाल्स समीकरण को लिखा जा सकता है:

Accepted Answer

संपीडन गुणांक (Z)=RTPV​एक मोल वास्तविक गैस के लिए वांडर वाल समीकरण(P+V2a​)(V−b)=RTनिम्न दाब पर, आयतन बहुत अधिक होगा। अतः अत्यधिक आयतन V की तुलना में b को नगण्य मानने पर, अतःV−b≈V(P+V2a​)V=RTPV+Va​=RTPV=RT−Va​RTPV​=1−VRTa​अत: Z=1−VRTa​

Question 2

Which one, among the following, is the Van der Waals' equation, describing the behaviour of one mole of a real gas over wide ranges of temperature and pressure?

Accepted Answer

The van der Waals' equation for n moles of a gas is[p+V2n2a​](V−nb)=nRT
For one mole (n=1)(p+V2a​)(V−b)=RT

Question 3

The unit of van der Waal's constant ′b′ is:

Accepted Answer

Since b corresponds to the total volume per mole occupied by gas molecules, it closely corresponds to the volume per mole of the liquid state, whose molecules are closely layered. Hence, all are the units of van der Waal's constant b.

Question 4

The van der Waals' equation for real gas is:

Accepted Answer

The van der Waals' equation for n moles of real gas is:(P+V2n2a​)(V−nb)=nRT...(1) It can also be written as: P=V−nbnRT​−V2an2​ For 1 mole of real gas, it can also be written as:(P+V2a​)(V−b)=RT (By Putting n=1 in equation (1)) Hence, all are correct.

Question 5

The unit of van der Waals' constant ′a′ is:

Accepted Answer

a=P×V2= atm litre 2 mol−2= dyne
cm4 mol−2=Newton m4 mol−2=atm dm6 mol−2

Question 6

Volume occupied by molecules of one mole gas at NTP, each having radius of 10−8 cm is:

Accepted Answer

Volume of molecules in one mole
​=4×N×V=4×N×34​πr3=4×6.023×1023×34​×722​×(10−8)3=10.09 mL​

Question 7

At relatively high pressure, van der Waals' equation reduces to:

Accepted Answer

Van der waal equation is,
(P+V2an2​)(V−nb)=nRT…. for n moles of gas
For 1 mole, the equation becomes,
(P+V2a​)(V−b)=RT
At high pressure, the Pcorrection ​= negligible but volume correction cannot be regarded as negligible.
​P(V−b)=RT or PV−Pb=RT or PV=RT+Pb​

Question 8

In van der Waals’ equation of state of the gas, the constant ′?′ is a measure of:

Accepted Answer

In van der Waals equation of state, the constant b represent excluded volume and are specific to a particular gas. The constant b is a measure of the volume occupied by the molecules.

Question 9

A real gas at high pressure occupies............ under identical conditions:

Accepted Answer

When the pressure is high, volume would be small. So, exclude volume ' b ' cannot be ignored. But as the pressure is high, the quantity V2a​ may be neglected in comparison with P. So the van der Waals' equation reduces to
P(V−b)=RT
PV−Pb=RT V=(PRT​+b)​ i.e., Vreal ​=Videal ​+b∴Vreal ​>Videal ​​

Question 10

The numerical value of ' a ' the van der Waals' constant is maximum for:

Accepted Answer

Van der Waal's constant a depend upon attraction.
a ∝ attraction
In the given option there is a maximum attraction in NH3​ because there is hydrogen bonding and in rest, all have Van der Waal's force.

Question 11

For one mole of a van der Waals gas, the compressibility factor Z(=RTPV​) at a fixed volume will certainly decrease if: [Given: ′a′,′b′ are standard parameters for van der Waals gas]

Accepted Answer

Compressibility factor Z=mPV​
Z=RTPVm​​=(Vm​−bVm​​)−(RTVm​a​)
=1−Vm​b​1​−RTVm​a​...(i)Considering the equation (i) the value of compressibility factor Z decreases as the value of ' b ' decreases and the value of compressibility factor Z decreases as the value of ' a ' increases at constant temperature and fixed volume.

Question 12

The behavior of a real gas is usually depicted by plotting compressibility factor Z versus p at a constant temperature. At high temperature and high pressure, Z is usually more than one. This fact can be explained by van der Waals equation when:

Accepted Answer

Van der Waal's equation is given by
(p+V2n2a​)(V−nb)=nRT
At high pressures, b cannot be ignored as the volume of the gas is very low. At high temperatures a can be ignored.Therefore,
p(V−b)pVRTpV​Z​=RT=RT+pb=RTRT​+RTpb​=1+RTpb​​

Question 13

Using van der Waals equation, calculate the constant, a (atm L2 mol−2) when two moles of a gas confined in a four liter flask exerts a pressure of 11 atm at a temperature of 300 K. The value of b is 0.05 L mol−1.

Accepted Answer

Van der Waals equation is given by
(p+V2n2a​)(V−nb)=nRT...(i)Substituting V=4 L,p=11 atm, T=300 K,b=0.05 L mol−1 and n=2 in Eq. (i), we get
​(11+(4)2(2)2×a​)(4−2×0.05)=2×0.082×300⇒a=6.5 atm L2 mol−2​

Question 14

Compressibility factor for a gas is represented as Z=5P−T3000P​. If the Boyle's temperature of gas is x K. Then the value of 10x​ is (in K):

Accepted Answer

At Boyle's temperature, dPdZ​=0  ⇒5−T3000​=05=T3000​T=600 K.  Hence, the value of 10x​ is 60K.

Question 15

In van der Waals' equation of state of the gas law, the constant ' b ' is a measure of:

Accepted Answer

Van der Waal's equation of state is (P+V2an2​)(V−nb)=nRTIn the ideal gas equation, we neglect the intermolecular attractions and volume occupied by molecules. However, the real gas equation accounts for both of these. The parameter a is a measure of intermolecular attraction and the parameter b is a measure of the volume occupied by the molecules. In particular, if the volume of a single gas molecule is v, then b=4vNA​. Hence b is four times the molar volume of the gas. This b gives the effective size of the gas molecule.

Question 16

The ratio a/b (the terms used in van der Waals' equation) has the unit:

Accepted Answer

From van der Waal's equation[P+V2a n2​][V−nb]=nRTunits of 'a' are:  atm litre2 mol−2=atm dm6 mol−2= dyne cm4 mol−2The units of 'b' are: litre mol−1=dm3 mol−1=cm3 mol−1So their ratio a/b is atm. lit/mol=atm dm3/mol=dyne cm/mol

Question 17

The term that corrects for the attractive forces present in a real gas in the Van der Waals' equation is:

Accepted Answer

In the van der Waals' equation(p+V2an2​)(V−nb)=nRTThe additional factor in pressure, i.e. V2an2​ corrects for attractive forces while b corrects for molecular volume.

Question 18

At low pressure, the Vander Waal's equation is reduced to:

Accepted Answer

Vander waal equation is given as(P+Vm2​a​)(Vm​−b)=RTAt low pressure, Volume (Vm​) will be sufficiently large and b may be ignored in comparison.Then it reduced to, (P+Vm2​a​)(Vm​)=RTPVm​+Vm​a​=RTPVm​=RT−Vm​a​⇒Z=RTPVm​​=1−RTVm​a​.

Question 19

At lower temperatures, all gases except H2​ and He show:

Accepted Answer

For H2​,He , Z>1 due to repulsive forces.

Question 20

The pressure of 2 moles of ammonia at 27∘C when its volume is 5 L according to van der Waals' equation is: (Given, a=4.17 atm L2 mol−2,b=0.03711 L mol−1)

Accepted Answer

T=27+273=300 KR=0.0821 L atm K−1 mol−1n=2 mol,  a=4.17,b=0.03711, V=5 L
(P+V2an2​)(V−nb)=nRT(P+524.17×22​)(5−2×0.03711)=2×0.0821×300
(P+254.17×4​)(5−0.07422)=600×0.0821(P+0.67)(4.92578)=600×0.0821P+0.67=4.92578600×0.0821​P+0.67=10.00P=10.00−0.67=9.33 atm.

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Solution for Class 11, Chemistry, States of matter

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