Thermodynamics Questions
Question 1
Two thermally insulated vessels are filled with equal number of moles of air and connected by short tube with a value. If P, V and T and pressure, volume and temperature of gas in first vessel and 2P, V and T / 2 be pressure, volume and temperature of gas second vessel then find temperature and pressure of air after opening the value.
Solution
On opening of valve air at pressure P and total volume of air in both the vessels would be 2V. since number of moles of air in both the vessels is equal so. from ideal gas equation we have
for vessel
(1) PV = μ_{1}RT
(2) 2PV = μ_{2}RT/2
after air is mixed up
2P^{'}V =(μ_{1}+ μ_{2})RT^{'}
Where P^{'} and T^{'} are final pressure and temperature
Again considering two portions of air as single system. Since the system is a thermally insulated vessel containing air so total heat transfer of combined system is zero i.e. Q = 0
Since no mechanical work is done on the system or done by the system => work done W = 0.
As no heat transferred takes place and no net amount of work has been done so there would be no change in internal energy of the combined system and would equal sum of internal energies of two partitions of air.
U_{i} = U_{1}+ U_{2}
= μ_{1}RT/(γ1)+μ_{1}R(T/2)/(γ1)
and final internal energy
U_{f} = (μ_{1}+ μ_{2})RT^{'}/(γ 1)
μ_{1}RT/(γ1) + μ_{1}R(T/2)/(γ1) = (μ_{1}+ μ_{2})RT^{'}/(γ 1)
or T^{'} = (μ_{1}+ μ_{2}) = (2μ_{1}+ μ_{2})T/2
μ_{1}= PV/RT
μ_{2}= 4PV/RT
=> T^{'}[(PV/RT + 4PV/RT) = 1/2(2PV/RT + 4PV/RT)T
5T^{'} = (1/2) 6T
T^{'} = (3/5 )T
P^{'} = (μ_{1}+ μ_{2})RT^{'}/2V
=5PV/RT x R/2V x 3/5 T
P^{'} = (3/2)P
So final answer is
T^{'} = (3/5 )T
P^{'} = (3/2)P
Question 2
Calculate the heat absorbed by the system in going through one cycle for the cyclic process
shown in the Figure
Solution
Since the process is cyclic so change in internal energy would be zero.
From first law of thermodynamics
W = Q
now work done is equal to area under this cycle which is
$W = \pi R^2$
$= \pi (100 cc)(100 kpa)$
$=3.14 \times 10$
$= 31.4$ J
=> heat absorbed by system during cyclic process is 31.4 J.
Question 3when water is boiled under a pressure of 2 atm, the heat of vaporization is $2.20 \times 10^6$ J Kg
^{1} and the boiling point is 120 °C. At this pressure 1 Kg of water has a volume $10_{3}$ m
^{3}, and 1 kg of steam a volume of 0.824 m
^{3},
(a) Compute the work done when 1 kg of steam is formed at this temperature
(b) Compute the increase in internal energy
Solution
(a) Work done in forming 1 Kg of steam at 120 °C is
$W = P dV$
= 2 x 10^{5} x 0.824m^{3} x 1.01
= 1.67 x 10^{5} J.
(b) Increase in internal energy is
$\Delta U = Q  W$
where Q = ML_{v}
where L_{v} is latent heat of vaporization of water
Q = 1 kg x 2256 J / gm
= 1000 x 2256 J / gm
= 2.256 x 10^{6} J
now $ \Delta U = 2.256 \times 10^6  .167 \times 10^6$ J
$= 2.092 \times 10^6 $ J
Question 4
Figure below shows different paths. connecting state i to state f in PV diagram along which gas can be taken Rank paths according to
(a) Change in internal energy
(b) Work done by gas
(c) The magnitude of energy transferred as heat.
Solution
(a) Change in internal energy ΔU remains same for all paths as it depends on initial and final points i and f not on the paths through which system is taken from initial point to final point
(b) Work done by gas along these paths is ranked as 1, 2, 3, 4 i.e., work done is greatest along path 1 because area under path 1 is greatest and area covered under path 4 is lowest
(c) from first law of thermodynamic
ΔU = Q  W
or Q = ΔU + W
from this magnitude of energy transferred as heat is greatest along path 1 and then subsequently on path 2,3 and 4.
Question 5
Two moles of helium (γ=5/3) are initially at a temperature of 27 °C and occupy a volume of 20L. The helium is first expanded at constant pressure until the volume has tripled, and then adiabatically until the temperature returns to its initial state. given for helium
C
_{p} = 20.78 J/molk
(a) Draw diagram of process in PV plane
(b) What is the total heat supplied in the process
(c) What is the total change in internal energy of helium.
(d) What is the total work done by the helium.
(e) What is the final volume and pressure.
Solution
(a) PV diagram of process is
(b) since the gas is expending at constant pressure so when its volume gets tripled its temperature also gets tripled.
for gas expending at constant pressure heat dQ flowing into gas is
dQ= nC_{p}dT
Integrating both sides with limits of T going from T_{1} and T_{2}
∫dQ=∫nC_{p}dT
Q=nC_{p} T_{2}  T_{1})
given that
C_{p} = 20.78 J/molk
n = 2 mol.
T_{1} = 27°C = 300 k
T_{2} = 3 x 300 = 900 k
= 2 x 20.78 x (900  300)
= 2.5 x 10^{4} J
(c) Total change in internal energy of helium is zero as initial and final temperature of the process is same
(d) for total work done by helium calculate work done by gas in isobaric process AB and also work done by gas during adiabatic process BC.
now from ideal gas equation
PV = μRT
Initial pressure
P = μRT/V = (2 x 8.314 x 300)/(20 x 10^{3})
= 2.49 x 10^{5} N/m^{2}
At point B
P_{B} = P_{A} = 2.49 x 10^{5} N/m^{2}
volume V_{B} = 3V_{A} = 60 x 10^{3} m^{3}
and T_{B} = 900 k.
Thus work done by gas during isobaric process AB
= 2.49 x 10^{5} (60  20) x 10^{3}
=9960 J.
Work done by gas during adiabatic process BC
= (μR/(1  γ))(T_{B}  T_{C})
= [2 x 8.314/1 (5/3)] [30900]
;= 14940
Therefore net work done by gas
W = W_{AB} + W_{BC}
= 9960 J + 14940 J
= 2.49 x 10^{4} J
(e) from adiabatic equation
TV^{γ1} = Constant
T_{B}V_{B}^{γ1} = T_{C}V_{C}^{γ1}
or, (V_{C} / V_{B})^{γ1} = T_{B}/T_{C} = 900 / 300 = 3
V_{C}/V_{B} = (3) ^{1/γ1} = (3) ^{1.5}
= 5.196
V_{C} = 5.196 x V_{B}
= 5.196 x 60 x 10^{3}m^{3}
= 311.7 x 10^{3}m^{3}
final pressure
P_{C} = μRT_{C} / V_{C}
= 2 x 8.3 x 300 / 311.7 x 10^{3}
= .159 x 10^{5} N/m^{3}
Question 6
Temperature of a system containing 1 mole ideal gas increases by on amount ΔT = 30 K as a result of heating at constant pressure. If gas obtains an amount of heat Q= 3.2 KJ. then calculate
(a) work done by the gas
(b) change in its internal energy
(c) value of γ= C
_{p} / C
_{v}
Solution
(a) Work done when system undergoes isobaric process is
W = PdV
from ideal gas equation
PV = μRT
here μ = 1
=> PdV = RμT
or W = RΔT
, = 8.314 x 80
= .665 KJ
(b) from first law of thermodynamics
ΔU = Q W = Q RΔT
given that Q= 3.2 KJ
ΔU = 3.2 KJ  .665 KJ
= 2.536 KJ
(C) Again the know that
ΔU = μC_{v}ΔT
= RΔT / γ1
γ1 = RΔT / ΔU
or γ= (RΔTT / ΔU) +1
= (.665 KJ x 8.314/ 2.535KJ) +1
= 1.26
Question 7
Two moles of a certain ideal gas at a temperature T
_{0} = 300K were cooled isochorically so that the gas pressure reduced two times. Then as a result of the isobaric process the gas expanded till its temperature get back to the initial value.
(a) What would be total change in internal energy.
(b) Total work done by the system.
(c) Find the total amount of heat absorbed by gas in this process
Solution
(a) Change in internal energy ΔU = 0 for the whole process
(b) during Isochoric process work done would be equal to zero = W_{1} = 0
now work done during isobaric process is
W = PΔV
= μRΔT
At the end of Isochoric process temperature of the gas can be calculated as follows
In the beginning of Isochoric process ideal gas equation is
PV=nRT_{0} (1)
During Isochoric process volume remains constant and at the end of the process pressure becomes P/2 and temperature becomes T'
Thus at the end of process ideal gas equation becomes
PV=nRT' (2)
From 1 and 2 T'=T_{0}/2
Now since temperature is increasing from T_{0}/2 to T_{0}
W=μR(T_{0} (T_{0} /2))
(c) total heat absorbed is given by ΔU = QW
from first law of thermodynamics
Q= W
total heat absorbed
Q= μRT_{0} (11/n)
= 2 x 8.314 x 300 x (11 / 2)
= 2494.25
= 2.5 KJ
Question 8
An ideal gas where adiabatic exponent is equal to γ, is expanded so that the amount of heat transferred to the gas is equal to the decrease of internal energy find
(a) molor heat capacity of gas in this process
(b) equation of process in variables T, V
(c) Work performed by one mole of gas when its volume increases by η time if T
_{0} is the initial temperature of the gas.
Solution
(a) As given in the question
$Q= \Delta U$
so $dQ= dU=  \mu C_v dT$
$= \frac {\mu R}{\gamma 1} dT$
now molar heat capacity
$C = (\frac {1}{\mu}) (\frac {dQ}{dT})$
$=  \frac {R }{ \gamma 1}$
(b) From first law of thermodynamics we have
$dQ= dU + dW$
but here : dQ= dU
$ dU= dU+ dW$
or $ 2 dU = dW$
or $2 \mu C_v dT = P dV$
or $(2\ mu \frac {R }{ \gamma 1} dT = (\frac {\mu RT }{ V}) dV$
$(\frac {dT}{ T}) + (\frac {\gamma 1 }{ 2 V})(\frac {dV}{V})=0$
On integrating me get
$TV^{\frac {\gamma 1}{2}} = constant$
which is the required equation
(c) As found earlier work done
$dW =  2dU$
or $W =  2 (U_2  U_1 )$
$= 2 C_vT_0((\frac {T }{ T_0})  1)$
= 2 C_{V} T_{0} (1  (T / T_{0}))
Now as we know that
TV^{(γ1)/2} = constant = T_{0}V_{0}^{(γ1)/2}
(T / T_{0}) = (V_{0} / V) ^{(γ1)/2} = 1(1 / η^{(γ1)/2})
So, W = 2 Cv T_{0} (1  T / T_{0})
=2 C_{V} T_{0} [1  1 / (η)^{(γ1)/2}]
= (2RT_{0} / γ1) [1  1 / (η)^{(γ1)/2}]
Question 9An ideal gas is taken from an initial state i to a final state f in such a way that the ratio of the pressure to the absolute temperature remains constant. What will be the work done by the gas.
Solution
Given that
P / T = const.
from ideal gas equation
P / T = μR / V = constant
V = μRT / P = constant
Since volume remains constant in this process =>V = 0
or W = pdV = 0
Work done during the process is zero.
Question 10 Consider the cyclic process ABCA, shown in the Figure, performed on a sample of 2.0 mole of an ideal gas. A total of 1200 J of heat is withdrawn from the sample as the process. Find the work done by the gas during the part BC.
Solution
Given that number of moles of gas is μ= 2.0 mole and heat rejected Q= 1200 J.
given process is cyclic so dU= 0
from first law of thermodynamic
dQ= dW
during CA volume remains constant
W_{CA} = 0
along AB
T ∝ V
=> T /V = const.
now from ideal gas eqn.
PV = μRT
P=μRT /V=constant
W_{AB} = P(V_{2}  V_{1})
= μRT_{2} μRT_{1}
= 2 x 8.3 x (500  300)
= 3320 J
now total work done
W = W_{AB} + W_{BC}
W = 3320 + W_{BC}
from first law of thermodynamics
Q= W
AS heat is being taken of
Q = W
 1200 J = 3320 J + W_{BC} or W_{BC} =  4520 J
Question 11 Consider a cylindrical tube of volume V with adiabatic walls containing an ideal gas. The internal energy of this ideal gas is given by 1.5 μRT. The tube is divided into two equal parts by a fixed diathenic wall. Initially, the pressure and the temperature are P
_{1}, T
_{1} on the left side and P
_{2}, T
_{2} right. The system is left for sufficient time so that the temperature becomes equal on the two sides.
(a) How much work has been done on the left part.
(b) Find the final temperature on the two sides.
(c) Find the final pressure on the two sides.
Solution
(a) W = 0 since vessel has fixed volume and no heat is exchanged between system and surrounding total internal energy of system will remain constant.
(b) Total energy of system before and after the process will remain some now before process internal energy of system is
U_{i}= 1.5 μ_{1} RT_{1} + 1.5 μ_{2} RT_{2}
and after the process
U_{f}= 1.5 (μ_{1} + μ_{2}) RT
where T is the final equilibrium temperature
Now U_{i}=U_{f}
1.5 μ_{1} RT_{1} + 1.5 μ_{2} RT_{2}= 1.5 (μ_{1} + μ_{2}) RT
or T = (μ_{1}T_{1} + μ_{2}T_{2}) / (μ_{1} + μ_{2}
) from ideal gas equation
μ= PV / RT
=> μ_{1} = P_{1}V_{1} / RT_{1}
=> μ_{2} =P_{2}V_{2} / RT_{2}
So
T = (P_{1} + P_{2}) T_{1}T_{2} / (P_{1}T_{2} + P_{2}T_{1})
or T = (P_{1} + P_{2}) T_{1}T_{2} /λ
where λ = P_{1}T_{2} + P_{2}T_{1}
c) If P_{1}^{'} and P_{2}^{'} are pressures on left and right sides respectively then
P_{1}V_{1}/ RT_{1} =P_{1}^{'}V_{1} / RT
=>
P_{1}^{'} = P_{1}T / T_{1}
putting value for T
= P_{1} (P_{1} + P_{2}) T_{1}T_{2} /λ T_{1}
or P_{1}^{'} = T_{2}P_{1} (P_{1} + P_{2}) T_{1}T_{2} /λ
similarly
P_{2}^{'} = P_{2}T / T_{2}
P_{2}^{'} = T_{1}P_{2} (P_{1} + P_{2}) T_{1}T_{2} /λ
Question 121 moles of an ideal gas whose adiabatic exponent equals y undergoes a process in which gas pressure relates to the temperature as
P=aT
^{2} where a is constant.
If the temperature get an increment of ΔT,find following
a. Change in Internal energy
b. Work performed by the gas
c. Molar heat capacity of the gas in the process
d. find the amount heat supplied
Solution
a. Change in internal energy = C_{v}ΔT=RΔT/(y1)
b. work done by the gas
given that
P=aT^{2}
Now we know that PV=RT
so, P=a(PV/R)^{2}
PV^{2}=R^{2}/a
PV^{2}=Constant
Now we know that
for PV^{y}=Constant
Work done is given as=RΔT/(1y)
Similarly for PV^{n}=Constant
Work done =RΔT/(1n)
Therefore work done in this case
W=RΔT/(12)
=RΔT
c. Molar heat capacity is given by
$C= \frac {dQ}{dT}$
$C= \frac {C_v dT + P dV}{dT}$
C= C
_{v} + P(dV/dT)
=C
_{v} + (RT/V)(dV/dT)
Now P=aT
^{2}
RT=aVT
^{2} as PV=RT
VT=constant
TdV + VdT=0
dV/dT=V/T
So C= C
_{v} + (RT/V)(V/T)
=C
_{v} R
=[{R/(y1)} R]
d. Q= ΔU +W
= [{RΔT/(y1)} RΔT]
=ΔT[R/(y1) R]
Question 13Two vessels A & B of equal volumes V
_{0} are connected by the narrow tube which can be closed by the valve. The vessels are fitted with the piston which can be moved to change the volumes.Initially the valve is open and vessel contains ideal gas whose adiabatic exponents is Y at atmosphere pressure P
_{0} and atmosphere temperature T
_{0}.The walls of the vessel A are adiabatic while the walls of the vessels B are diathermic. The valve is now closed and piston is slowly moved to triple the original volume.
a. Find the temperature and pressure in the two vessels.
b The valve is now open for sufficient time so that gases acquire a common temperature and pressure. Find the new values of pressure and temperature
Solution
a. for Vessel A ,walls are adiabatic
so PV^{γ}=constant
P_{0}V_{0}^{γ}= P(3V_{0})^{γ}
P= P_{0}/3^{γ}
Also
TV^{γ1}=constant
T= T_{0}/3^{γ1}
For vessel B ,wall are diathermic, so final temperature is going to be T_{0} only
So P_{0}V_{0}=P* 3V_{0}
P=P_{0}/3
b. Now when the valve is opened again, final temperature will be again T_{0}
Also
Initially P_{0}V_{0}=nRT_{0}
finally
P*3V_{0}=nRT_{0}
therefore
P=P_{0}/3
Question 141 mole of an ideal gas whose adiabatic exponents is γ is enclosed in the vertical adiabatic vessel fitted with moving frictionless piston whose weight is W and crosssectional area is A.The atmospheric pressure is P
_{0}.Initial volume is V
_{0}.Heat is supplied to the gas to double its volume.
a. Find the initial and final state of the gas
b. Find the work done by the gas
C. Find the change in internal energy
Solution
a.
Initial State
Initial Pressure = [P_{0} + (W/A)]
Initial volume = V_{0}
Initial temperature from PV=RT
=[P_{0} + (W/A)]V_{0}/R
Final state
Pressure remain constant so
Final Pressure = [P_{0} + (W/A)]
Final volume = 2V_{0}
Final temperature from PV=RT
=2[P_{0} + (W/A)]V_{0}/R
b. Work done by the gas
= ∫PdV
Since pressure remains constant
=[P_{0} + (W/A)]V_{0}
c. Change in internal energy is given by
= C_{V}ΔT
= [{R/(γ1)})[P_{0} + (W/A)]V_{0}/R
= [P_{0} + (W/A)][V_{0}/(γ1)]
d Heat supplied
Q= ΔU +W
= [P_{0} + (W/A)][V_{0}/(γ1)] + [P_{0} + (W/A)]V_{0}
=[P_{0} + (W/A)][V_{0}y/(γ1)]
Question 15Two samples A and B whose adiabatic exponents is γ are initially kept in the same state (P
_{0},V
_{0},T
_{0}).Now sample A is expanded such that volume becomes 2V
_{0} and Q=0 for the process. Sample B is expanded such that volume becomes 2V
_{0} and ΔU=0 for the process
a. Find the ratio of the final pressure of the two samples
b. Find the ratio of the final temperature of the two samples
c. Find the ratio of the work done by the two samples
Solution
a. For sample A
Since Q=0 therefore Adiabatic process
P_{0}V_{0}^{γ}= P_{A}(2V_{0})^{γ}
P_{A}=P_{0}/2^{γ}
For sample B
Since ΔU=0,therefore Isothermal
So P_{0}V_{0}=P_{B} 2V_{0}
P_{B}=P_{0}/2
So P_{A} : P_{B} = 1 : 2^{γ1}
b. For sample A
T_{0}V_{0}^{γ1}=T_{A}(2V_{0})^{γ1}
T_{A} = T_{0}/2^{γ1}
for sample B
T_{B} = T_{0}
So T_{A} : T_{B} = 1 : 2^{γ1}
c. Work done by the Sample A
$W=\frac {P_iV_i P_fV_f}{\gamma 1}$
$=\frac {P_0V_0  (\frac {P_0}{2^{\gamma} }) \times 2V_0}{\gamma 1}$
$=\frac {P_0V_0 (2^{\gamma 1} 1)}{2^{\gamma 1}(\gamma 1)}$
Work done by the Sample B
W
_{B}= RT
_{0}ln2
= P
_{0}V
_{0}ln2
Now ratio
W
_{A} : W
_{B}= 12
^{1y} : (y1)ln2
Question 16One mole of an ideal monatomic gas is taken round the cyclic process ABCDA as shown in figure.
a. Work done by the gas
b Heat absorbed by the gas in AB and BC
c. Heat in process CD
d. Find the temperature at C and D
e. Maximum temperature attained by the gas during the cycle
f. Net change in the internal energy and the heat
Solution
For monatomic gas
C_{p}=5R/2
C_{v}=3R/2
a Work done by the gas = Area enclosed by the curve ABCDA
=3P_{0}V_{0}
b. Heat absorbed the gas in AB
Q_{AB}=C_{v}(T_{B}T_{A})
=3R/2(3P_{0}V_{0}/R P_{0}V_{0}/R)
=3P_{0}V_{0}
Heat absorbed the gas in BC
Q_{BC}=C_{P}(T_{C}T_{B})
=5R/2(6P_{0}V_{0}/R  3P_{0}V_{0}/R)
=15P_{0}V_{0}/2
c. Heat rejected in DA
Q_{DA}=C_{P}(T_{A}T_{D})
=5P_{0}V_{0}
Now for the cycle process
Q_{AB}+Q_{BC}+Q_{CD}+Q_{DA}=W
So
Q_{CD}= 5P_{0}V_{0}/2
d. from diagram and PV=RT
T_{C}=6P_{0}V_{0}/R
T_{D}=3P_{0}V_{0}/R
e. Max temperature will be on the slope CD
Equation of Slope CD as Coordinated system
P=(2P_{0}/V_{0})V + 7P_{0}
Now PV=RT
So
RT= (2P_{0}/V_{0})V^{2} + 7P_{0}
For max
dT/dV should be zero
So RdT/dV=4P_{0}V/V_{0} +7
V=7V_{0}/4
T_{max}= 49P_{0}V_{0}/4R
f. ΔU=0
Net heat=3P_{0}V_{0}
Question 17
An ice cube of mass .1kg at 0 °C is placed in an isolated container which is at 227° C. The specific heat of the container varies with temperature according to the empirical formula
S=A + BT
where A is 100cal/kgK and B is 2x10
^{2} cal/kgK
^{2}
if the final temperature of the container is 27°C.Determine the mass of the container.
Specific heat of water is 10
^{3} cal/kgK
Latent heat of fusion of water is 8x10
^{4} cal/kg
Solution
Let m be the mass of the container. Since specific heat varies with temperature, we will have to integrate to total heat lost from container.
dQ=msdT
dQ=m(A+BT)dT
Integrating for the upper and inner limit 300 and 500 respectively
Q= ∫m(A+BT)dT
Q=m[AT +BT^{2}/2]
=21600m
now heat gained by the ice
Q= heat given to 0° (ice) to 0°(water) + heat required to raise the temperature from o to 27°C
Q=mL + mcΔT
=.1 * 80000 + .1*10^{3}*27
=10700 cal
Now heat gained =Heat lost
So
21600m=10700
m=.495kg
Question 18A gaseous mixture is such that
Gas

No of moles

Adiabatic coefficient

A

1

1.67

B

2

1.4

The gaseous mixture is taken through an adiabatic process where temperature of the mixture drops by ΔT.
a. Find the work done by the mixture
b. If m
_{1} and m
_{2} are the relative molecular mass of the gas A and B respectively. Find the speed of sound in the mixture at Temperature T
_{0}
Solution
a.Let C_{p} ,C_{v} and y_{m} are the heat capacities and adiabatic coefficient of the mixture.
then y_{m}=C_{p}/C_{v}
Now C_{p}=n_{A}C_{pA}+n_{B}C_{pB}/n_{A}+n_{B}
Also C_{v}=n_{A}C_{vA}+n_{B}C_{vB}/n_{A}+n_{B}
So y_{m}=n_{A}C_{pA}+n_{B}C_{pB}/n_{A}C_{vA}+n_{B}C_{vB}
Substituting the values
y_{m} =1*5/2R + 2*7/2R/1*3/2R +2*5/2R
=19/13
Work done by the mixture is given as
=nRΔT/y1
=39RΔT/6
b. Molar mass of the mixture will be =m_{1}+2m_{2}
Speed of sound=√yRT/M
=&radic 19RT_{0}/13(m_{1}+2m_{2})
Question 19
A diatomic gas whose adiabatic coefficient is γ=1.4 is taken through XYZAX cycle.
X>Y Adiabatic compression
Y>Z Isobaric expansion
Z>A Adiabatic expansion
A>X Isochoric process
find the following
 Molar heat capacities in each process
 if the volume ratio is V_{X}/V_{Y}=16 and V_{Z}/V_{Y}=2 and T_{Y}=636° C
a. find the heat absorbed in the process Y>Z
b. find the net work done by the gas in the cycle
c. find the ratio of pressure P
_{Y}/P
_{X} and P
_{Z}/P
_{A}
Solution
n=1
γ=1.4
1. Molar heat capacity of the gas in the Process X>Y (Adiabatic)=0 as Heat is zero
Molar heat capacity of the gas in the Process Y>Z (pressure constant)=C_{p} =7/2R
Molar heat capacity of the gas in the Process Z>A (Adiabatic)=0 as Heat is zero
Molar heat capacity of the gas in the Process A>X (Volume constant)=C_{v} =5/2R
2. V_{X}/V_{Y}=16
For adiabatic compression X>Y
T_{X}V_{X}^{γ1}=T_{Y}V_{Y}^{γ1}
T_{X}= T_{Y}(V_{Y}/V_{X})^{γ1}
=909 (1/16)^{.4}
=300K
Now Y>Z ,According to Charles law
V_{Z}/T_{Z}=V_{Y}/T_{Y}
So T_{Z}= 909*2=1818 K
Heat absorbed in Y>Z>
=nC_{p}(T_{Z}T_{Y})
=.3182R
W_{XY}= nR(T_{Y}T_{X})/γ1
=5R/2* 609
W_{YZ}=nR(T_{Z}T_{Y})
=909R
W_{ZA}= nR(T_{Z}T_{A})/γ1
Now we know that V_{X}/V_{Y}=16 and V_{Z}/V_{Y}=2
as V_{X}=V_{A}
So V_{Z}/V_{A}=1/8
Now for Z>A Adiabatic expansion
T_{Z}V_{Z}^{γ1}=T_{A}V_{A}^{γ1}
T_{A}=T_{Z}(V_{Z}/V_{A})^{γ1}
=791K
So
W_{ZA} =5R/2 * 1027
Also W_{AX}=0 as volume constant
Then
W_{NET}=909R + 5R/2(1027609)
=1954R
Now For adiabatic compression X>Y
P_{X}V_{X}^{γ}=P_{Y}V_{Y}^{γ}
P_{Y}/P_{X}= 16^{1.4}
Similarly
Now for Z>A Adiabatic expansion
P_{Z}V_{Z}^{γ}=P_{Z}V_{Z}^{γ}
P_{Z}/P_{A}= 8^{1.4}
Question 20
Consider one mole of perfect gas in a cylinder of unit cross section with a piston attached in below figure. A spring (spring constant k) is attached (unstrectched length L ) to the piston and to the bottom of the cylinder. Initially the spring is unstrectched and the gas is in equilibrium. A certain amount of heat Q is supplied to the gas causing an increase of volume from $V_0$ to $V_1$.
(a) What is the initial pressure of the system?
(b) What is the final pressure of the system?
(c) Using the first law of thermodynamics, write down a relation between Q, $P_a$, V, $V_0$ and k.
Solution
(a): It is considered that piston is mass less and piston is balanced by atmospheric pressure ($P_a$). So the initial pressure of system inside the cylinder = $P_a$.
(b) On supply heat Q. Volume of gas increases from $V_0$ to $V_1$ and spring stretched also.
$\Delta V = V_1  V_0$
If displacement of piston is x then volume increase in cylinder
$\Delta V= \text {Area of base} \times {height} = A \times x$
$A \times x = V_1  V_0 $
$x = \frac {V_1 V_0}{A}$
$x=V_1 V_0$
As A=1
Force exerted by spring $F_S = K(V_1 V_0 )$
So final total pressure on gas $P_f = P_a + K(V_1  V_0)$
(c) By Ist law of thermodynamics
$dQ = dU + dW$
$dU = C_V(T  T_0)$
$T = \frac {PV}{R} = \frac {[P_a + K(V_1  V_0)]V_1}{R}$
Now
W.D. by gas = p. dV + increase in PE of spring
$dW = P_a(V_1 V_0) + \frac {1}{2} K(V_1 V_0)^2$
Now $dQ = dU + dW$
So,
$Q= C_V(T  T_0) + P_a(V_1 V_0) + \frac {1}{2} K(V_1 V_0)^2$
Where $T= \frac {[P_a + K(V_1  V_0)]V_1}{R}$
It is required relation.
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