Everything that is really great and inspiring is created by the individual who can labour in freedom

– Albert Einstein

Equations to be remembered in kinetic theory of gases were discussed in the post dated 13^{th} March 2008, followed by a discussion of some typical multiple choice questions (for practice) in the post dated 17^{th} march 2008. You can access those posts by clicking on the

Important points related to thermodynamics were discussed in the post dated 21^{st} May 2008, followed by a discussion of some typical multiple choice questions (for practice) in the post dated 26^{th} May 2008. A couple of free response practice questions on thermodynamics were discussed in the post dated 3^{rd} June 2008. You can access them by clicking on the

Today we will discuss a few more multiple choice questions** **on Kinetic Theory and Thermodynamics.

(1) A region of the earth’s atmosphere contains *n* molecules (treated as ideal gas molecules) per unit volume. The temperature of air in the region is *T*. If *k* represents Boltzmann’s constant and *R* represents universal gas constant, the pressure of air in the region is

(a) *nT/k*

(b) *nkT*

(c) *RT/n*

(d) *nRT*

(e)* nRkT*

Since *PV =* *μRT** *where* **μ* is the number of moles in the volume *V* of the gas at the pressure *P* and temperature *T*,* *we have

*P = *(*μRT*)*/V*

Since *k = R/N* where *N* is Avogadro number, *R = Nk*.

Therefore, *P = *(*μNkT*)*/V*.

But* *(*μN*)/*V* = *n*, the number of molecules per unit volume so that *P = nkT*

[The expression for the pressure *P *of a gas according to the kinetic theory is

*P = *(1/3) *nmc*^{2}

The r.m.s.velocity *c *of the molecule is √(3*kT/m*) where *m* is the molecular mass. On substituting for *c *in the expression for* P *you can easily obtain the expression* P = nkT*].

(2) An ideal monoatomic gas is heated so that it expands at constant pressure. What percentage of the heat supplied to the gas is used to increase the *internal energy* of the gas?

(a) 100 %

(b) 0 %

(c) 60 %

(d) 50 %

(e) 40 %

Since a monoatomic gas molecule has 3 degrees of freedom, the molar heat capacity (also called molar specific heat) of the monoatomic gas at constant volume (*C*_{V}) is (3/2)*R* where *R* is universal gas constant. Its molar heat capacity at constant pressure (*C*_{P}) is (5/2)*R* . This follows from* **Meyer’s relation*, *C*_{P} = *C*_{V }+ *R*.

This means that when heat energy equal to (5/2)*R* joule is supplied to one mole of a monoatomic gas to make it expand at constant pressure, (3/2)*R* joule is used to increase the *R* joule is used for doing work against the forces which oppose the expansion. 3/2 is 60 % of 5/2. So the correct option is (c).

(3) Oxygen and nitrogen in two enclosures have the same mass, volume and pressure. The ratio of the

(a) ^{7}/_{8}

(b) ^{8}/_{7}

(c) 1

(d) ^{49}/_{64}

(e) ^{64}/_{49}

We have *PV = μRT* where* **μ* is the number of moles in the volume *V* of the gas at the pressure *P* and temperature *T*.

Since *P* and *V* are the same for the two gases,

*PV = μ*_{1}*RT*_{1}* = μ*_{2}*RT _{2} *where

*μ*

_{1}and

*μ*

_{2}are the numbers of moles of oxygen and nitrogen and

*T*

_{1 }and

*T*

_{2}are their temperatures respectively.

Therefore, *μ*_{1}*T*_{1}* = μ*_{2}*T _{2}* so that

*T*

_{1}/

*T*

_{2}=

*μ*

_{2}/

*μ*

_{1}

The samples of the two gases have the same mass (say, *m*).

Therefore, *μ*_{1} = *m/*0.032 and *μ*_{2} = *m/*0.028 since the molar masses of oxygen and nitrogen are 0.032 kg and 0.028 kg respectively.

Therefore,* T*_{1}/*T*_{2} = *μ*_{2}/*μ*_{1} = 0.032/0.028 = ** ^{8}/_{7}**.

(4) In a Carnot engine 7000 J of heat is absorbed from a source at 400 K and 4900 J of heat is rejected to the sink. The

(a) 200 K

(b) 220 K

(c) 260 K

(d) 280 K

(e) 340 K

We have *Q*_{1}/*T*_{1 }*= Q*_{2}/*T*_{2} with usual notations.

Substituting the given values, 7000/400 = 4900/*T*_{2}.

Therefore, *T*_{2} = (4900×400)/7000 = **280 K**

(5) Three moles of an ideal monoatomic gas is initially in the state A shown in the adjoining pressure-temperature graph. It is taken to state B without changing its pressure. If R is the universal gas constant, the work done by the gas in this process is

(a) 200 R

(b) 300 R

(c) 400 R

(d) 500 R

(e) 600 R

The work done by the gas in taking it from state A to state B = *P∆V* where *∆V* is the increase in volume at constant pressure *P*.

We have *PV* = *μ*R*T* where *μ* is the number of moles in the sample of the gas and R is the universal gas constant.

Therefore we have

*P∆V = μR ∆T* = 3×R(450 – 250) = 600R [Option (e)].

Hello, your web site is very useful for High School Physics Student.

ReplyDeleteI want to ask if the answer to question 3 is wrong?(the oxygen nitrogen question)

My answer is 7/8, thanks.

Alain (Hong Kong)

Hello Anonymous,

ReplyDeleteThe answer to question 3 is 8/7 itself.