Monday Jan 12:

 

Continue Review Ch 1-14 Final (Thurs 13 Jan)

 


 

  • R.M.S. speed of molecule: 

 Vrms = √(3RT/M) = √(3kBT/m).


The first one gives the r.m.s.speed in terms of molar mass M &

 

the universal gas constant R.

 

The second one gives the r.m.s. speed in terms of molecular

 

mass ‘m’ and Boltzman’s constant ‘kB.


 

Q#1:

 

Gases exert pressure on the walls of the container

 

because the gas molecules:
 

(a) collide one another
 

(b) exert intermolecular attraction
 

(c) possess momentum
 

(d) expand on absorbing heat
 

(e) exert repulsive force
 
 

 

 

 

 

Because of the momentum of the gas molecules, they collide

with the walls of the containing vessel and momentum

transfer takes place, resulting in a force on the walls. Pressure

is force per unit area. The basic reason for the pressure is the

momentum of the gas molecules [Option (c)]. 


 

Q#2:

 

A fixed mass of an ideal gas at pressure P is contained in

 

a closed vessel of volume V. It is heated so that the root

 

mean square velocity of the gas molecules is doubled.

 

The thermal expansion of the vessel is negligible. Then

 

the increase in pressure (∆P) of the gas is
 

(a) P
 

(b) 2P
 

(c) 3P
 

(d) 4P
 

(e) √2 P
 

 

 

 

 

 

 

 

 

Don’t pick out option (d) in a hurry.

 

The root mean square velocity of the gas molecules is

 

given by Vrms = √(3kT/m). The r.m.s. velocity is therefore

 

directly proportional to the square root of absolute

 

temperature. Since the r.m.s. velocity is doubled on heating

 

the gas, the temperature is quadrupled. By Charle’s law, we

 

have PV/T = constant for a given mass of gas. When the

 

temperature is quadrupled at constant volume, the

 

pressure must be quadrupled. The final pressure is thus

 

4P and the increase in pressure from 1P to 4P  is 3P.

 

 

Q#3:

 Equal number of oxygen molecules (molar mass m1)

and helium molecules (molar mass m2) are kept in two

identical vessels. If they are at the same temperature,

their pressures will be in the ratio

 

(a) m1/m2
 

(b) m2/m1
 

(c) √(m1/m2)
 

(d) √(m2/m1)
 

(e) 1
 

 

 

 

 

 

 

 

Most of you know that equal volumes of all gases

 

under the same conditions of temperature and

 

pressure contain the same number of molecules. If

 

equal numbers of molecules of different gases have the

 

same temperature and volume their pressure must

 

therefore be the same. The correct option is (e).
 

 

 

 

Q#4:
 

To decrease the volume of an ideal gas by 10% at

 

constant temperature, the pressure should be increased by


(a) 5% (b) 8.91% (c) 10% (d) 11.1% (e) 12.25%
 

 

 

 

 

 

 

 

 

Don’t pick out option (c) in a hurry.

 

We have P1V1 = P2×0.9V1 (constant temperature)

 

so that P2 = P1/0.9 --> 1.111P1

 

 The increase in pressure is 11.1% [option (d)].

 

 

 

Q#5:

 

When a gas contained in a closed vessel is heated

 

additional 1˚C, the pressure of the gas increases by 0.2%.


the final temperature of the gas is
 

(a) 200K (b) 361K (c) 500K (d) 501K (e) 601K
 

 

 

 

 

 

 

 

 

 

(PV/T)1 = (PV/T)2   volume is constant: i.e., V's are crossed out.

 

P/T = 1.002P/(T+1)  

 

1/T = 1.002/(T+1)   --> T + 1 = 1.002T

 

T = 500K. The final temperature is

 

T+1 = 501K.