Question

For this problem, let's assume instead that this transition is very sudden, as shown in the graph below. Heat capacity per molecule 100 K 800 K Temperature A. First, recall that heat capacity is defined as the amount of heat that must be added to raise the temperature of an object by a certain change in temperature: C = * = Q = CAT. a. How could you graphically determine how much heat was necessary to raise the temperature by a given AT, say from 2 K to 4 K?

Answer 1

A)

The heat capacity is given by

Heat capacity per molecule 100 K 800 K Temperature

a)

The heat required, = area under the graph between the temperature limits.

Q=CAT

where,

Q = heat required

C = heat capacity

А в Мја а1< 4K TOOK

Thus the area under the heat capacity - temperature curve gives the heat transferred

b)

Heat, Q = area ABCD - CXDT Area AB CD - 3 kg x (4-2) = 3 kB Ko = Boltzmann constant Kb = 1.38 x 1023 m2 kg 52,61 Heat required for one molecule = 3x1.38x10-23 Joules fore I mole Cimole 6.02 ax 1023 molecules) Q = 6.022x10 23 x 3 x1-38X10-23 Q = 24.93 Joules

B)

a)

Area AI Slaket- IL B A1 = Aa final temperature Tf>look. - Area Aa kel 0 AC ak 4k 80K TOOK 12010

From the graph, we can infer that,

since the mass is the same, for the same change in temperature,$\Delta T$,  gas with higher heat capacity value needs large heat transfer than the gas with low heat capacity value. In other words, when an equal mass of the gases are mixed,

the heat lost by gas at high temperature = heat gained by gas at low temperature

i.e. if the final temperature reached is Tf , then,

area under the graph between 80 K and Tf = area under the graph between Tf and 120 K

Since the area between 80 K and 100 K is smaller than area under 100K and 120K, the final temperature reached is not 100 K. Tf should be greater than 100 K

b)
graph is reproduced again

Area AI Slaket- IL B A1 = Aa final temperature Tf>look. - Area Aa kel 0 AC ak 4k 80K TOOK 12010

The temperature Tf will be greater than 100 K ( already explained ). It has been labeled in the figure.

c)

Simplified graph is drawn below

Asea ABCOEF = Heat gained by Sample A 3/2 kg- Area DGHE & Heat lost by o the sample. DE : line of Tf. TH FM too IC 12016 80K

d)

Heat gained, Q. CXOT since Both the mixtures are of equal moss, Heat gained by one molecule, = CXDT - Area ABCOEF Q = area ABIF + area COEI - 3 ICBX C100-800+ & KB x (Tf - 1000 KO [60 + SCTF-1009) Joules Total heat gained by sample = Q x 6.022x 1023

Heat lost by one molecule , = CXOTE area OG HE Pe - KBX Ciao-T4] Total heat lost by sample - Dex 6.022 x 1023. From inference, Total heat gained - Total heat lost 18 [ 60+ 5 CI+-100]] • Ske (120-14) 60+ S(Tf - 100) - S0120 - Tf] GO+ STF - 500 - 600 + STF = 0 10 Tf - 10400 Tf = 104 K

Thus the final temperature reached is 104 K

C)

GOK TF 10 - -- - - 12016 H

Now the area under the graph needs to be found out for obtaining heat. area under any curve, $f\left ( x \right )$ with the X axis is given by the formula,

ap (2) s ] = vacy

where, ' a ' and ' b ' are the x limits and area of the section between ' a ' and ' b ' is calculated.

Heat, Q- area under the graph - Scat Tf Heat gained, ag= Sc dT , where c= CCT) roic teat lost, an de= cdT 80

evaluate the integrals and equate the heat loss with the heat gain to find the final temperature.

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