MP203 Statistical and Thermal Physics. Solutions to Problem Set 3

Transcription

1 MP03 Statistica and Therma Physics Soutions to Probem Set 3 1. Consider a cyinder containing 1 mo of pure moecuar nitrogen (N, seaed off withamovabepiston,sothevoumemayvary. Thecyinderiskeptatatmospheric pressure and we can make accurate measurements of its voume. For the purpose of this probem, you may consider N to be an idea gas. (a One morning, we find that the voume of the nitrogen equas m 3 and its temperature is 7 C. What is the atmospheric pressure at the time of measurement? (b Later that day the temperature has increased to 17 C, whie the atmospheric pressure has remained unchanged. What is the voume of the nitrogen now? (c How much work did the nitrogen perform in its expansion? How much heat did it absorb in the process? (d Finay, we expand the gas adiabaticay (by puing the piston from the situation in part (b, unti it reaches its origina temperature of 7 C. i. How much work is done on the atmosphere in this process? ii. What is the fina voume of the gas? (a We use the idea gas aw in the form p nrt, with in this case n 1 mo, T T 1 7 C 80 K and m 3. This gives us p nrt (1mo 8.31J/(moK 80K m Pa. (1 (b Using the idea gas aw, we find that the voume at the time when the temperature T 17 C 90 K, is nrt p T T m m 3. ( (c The pressure was constant during the expansion, so the work performed is W p ( Pa ( m m 3 83J. (3 The heat absorbed can be cacuated using the first aw of thermodynamics. We have Q U + W, where U is the change in the energy of the gas. Here, we have a diatomic idea gas, so the energy is given by U 5 nrt. Hence U 5 nr T 5 (1mo(8.31J/(moK(10K 08J (4 Q U +W 08J+83J 91J. (5

2 (d In an adiabatic expansion, Q 0, so using the first aw, U W, where W is the work performed by the gas on the atmosphere. We now have a negative T 10 K, so we get W U 5 nr T 08J. (6 To find the fina voume, we need the formua for the adiabatics of an idea diatomic gas. We have (since dq 0 ( 5 Ti f i i du dw (7 5 nrdt pd nrt 5, 5 dt Tf T i T d dt T f i d d (8 (9 (10 5 n T f n f (11 T i i ( 5 Tf i n n (1 T i f T 5 i i T 5 f f (13 ( m m 3. (14 Note that in this particuar exam the reation (13 was not given, so the students woud have to either derive it as above, or remember it. Here, you are free to consut with your notes, so starting from (13 is sufficient. *. A monatomic idea gas is kept in a cyinder with a piston which aows the gas to be expanded or compressed. The gas is initiay at atmospheric pressure, p atm Pa and at room temperature, T 300K. (a The gas is adiabatiay expanded from an initia voume of m 3 to a fina voume of m 3. What is the fina temperature of the gas? (b We bring the expanded gas into contact with a box containing a voume box m 3 of air, which is intiay at temperature T i,box 300 K and at atmospheric pressure. We aow heat to fow from the air in the box to the expanded gas, keeping the expanded gas at a fixed voume. Eventuay the gas in the box and the gas in the cyinder have the same temperature. What is this fina temperature? (c We take the cyinder out of therma contact with the box and bring it into therma contact with the atmosphere. After a short time its temperature becomes equa to the ambient temperature of 300 K.

3 We now isothermicay compress the gas in the cyinder from its voume of m 3 back to its initia voume of m 3. How much work do we do in compressing the gas? How much heat does the gas reease into the atmosphere in the process? (a Using the reation between temperature and voume in an adiabatic expansion, for a monatomic gas with f 3 degrees of freedom, we have i T 3/ i f T 3/ f (15 ( /3 i ( 1 /3300K T f T i 189K. (16 f (b The heat reeased by the air is absorbed by the gas in the cyinder and both of these processes happen at constant voume. We can thus write C,box (T f T i,box C,cy (T f T i,cy, (17 where T i,cy is the temperature cacuated in part (a and T f is the joint fina temperature, and C,box,C,cy are the (temperature-independent heat capacities of the air in the box and gas in the cyinder respectivey. Here we used that the heat capacity of a (cassica idea gas does not depend on temperature. These are given by C,cy 3 N cyk B (monatomic gas, f 3 (18 C,box 5 N boxk B (diatomic gas, f 5 (19 We can find N cy and N box using the idea gas aw, N p. For the cyinder kt we can use the initia temperature, pressure and voume to give N cy ( ( Simiary, for the box, we find (0 N box ( ( ( We actuay ony need that box 8 cy at the same T and p, which gives us N box 8N cy. This gives us C,box 5 N boxk 0N cy k B 40 3 C,cy ( Substituting this in (17, we find that 40 3 C,cy(T f T i,box C,cy (T f T i,cy ( T f T i,cy T i,box (4 T f 3 43 T i,cy T i,box K+ 300K 9K. (

4 (c Since the temperature of the gas does not change, the energy U 3 NkT is constant and the first aw tes us that Q W, i.e. the heat absorbed by the gas is equa to the work done by it. This aso mean the heat Q reeased by the gas is equa to the work W we do in compressing it. We know that dw pd and hence the work done on the gas is W 1 0 pd 1 0 Nk B T d Nk B T n( 1 0 where 0 and 1 are the initia and fina voume of the compression process, so in this case 0 0m 3 and 1 10m 3. We aready cacuated in the soution to part (b that N cy (athough it is reay ony needed now. This then gives us W Q ( ( n( 1 J 30J 3. (a A quantity of an idea diatomic gas is kept in a cyinder with a piston which aows the gas to be expanded or compressed. Initiay, the gas has a voume itres and is kept at a temperature T h 350 C and pressure p Pa. The gas then undergoes an isotherma expansion to a voume 4 1. Compute the heat absorbed and the work done by the gas in this process. (b The gas then expands adiabaticay unti it reaches a temperature T 5 C. Compute the pressure p 3 and voume 3 of the gas at the end of this process, and the work done by the gas in the process. (c The gas is then compressed isothermay to a voume 4 3 /4, before undergoing an adiabatic compression to return to its origina state. Compute the tota work done by the gas in the entire cyce. (d The efficiency e of a cycic process is defined as e W Q in, where W is the net work done in the cyce and Q in is the heat absorbed in the high-temperature expansion (not the net heat absorbed. Cacuate the efficiency of the cycic process described above. (a In an isotherma expansion of an idea gas, we have p Nk b T constant, so the work done by the gas is given by In this case we get W 1 pd 1 p 1 1 d p 1 1 n 1. (6 W 1 p 1 1 n J J. (7

5 Since for an idea gas the interna energy depends ony on the temperature, there is no change in interna energy, so the heat absorbed equas the work done. (b In an adiabatic expansion, the work is equa to the change in interna energy U f Nk BT, where f 5 for a diatomic idea gas. Using the idea gas aw, we find for the work W done by the gas in this process W U f Nk B(T h T 5 Nk BT h ( 1 T T h 5 ( p T 5 155J ( 1 98K T h 63K The voume 3 at the end of this process is found from 0J. (8 T f/ h 3 T f/ ( f Th 3 T ( (9 98 The pressure can be found using p 3 p ( 3 γ Pa ( Pa. (30 (c We can determine the work done in the remaining haf of the cyce without any further significant cacuation. Firsty, the isotherma compression mirrors the first isotherma expansion, except that the temperature, and therefore the product p, is ower by a factor T /T h. We therefore have, for the work done by the gas at this stage, W 3 T W 1 T Q J 103J, (31 T h T h 63 where Q 1 is the heat absorbed in the first stage of the cyce. Secondy, in the fina adiabatic compression, the change in interna energy, and therefore the work done (since there is no heat, must exacty mirror that in the adiabatic expansion, since it simpy reverses the temperature change. Therefore W 4 W. Adding this up, we find that the tota work done by the gas is W W 1 +W +W 3 +W 4 W 1 +W 3 Q 1 T ( Q 1 1 T Q 1. (3 T h T h We wi ater see that this cycic process (caed a Carnot cyce has the maximum efficiency aowed by the aws of thermodynamics.