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/(γ-1)+μ_{1}R(T/2)/(γ-1)
and final internal energy
U_{f} = (μ_{1}+ μ_{2})RT^{'}/(γ -1)
μ_{1}RT/(γ-1) + μ_{1}R(T/2)/(γ-1) = (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
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.
(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
(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.
(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/mol-k
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)] [30-900]
;= 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^{γ-1} = Constant
T_{B}V_{B}^{γ-1} = T_{C}V_{C}^{γ-1}
or, (V_{C} / V_{B})^{γ-1} = T_{B}/T_{C} = 900 / 300 = 3
V_{C}/V_{B} = (3) ^{1/γ-1} = (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}
(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 / γ-1
γ-1 = RΔT / ΔU
or γ= (RΔTT / ΔU) +1
= (.665 KJ x 8.314/ 2.535KJ) +1
= 1.26
(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 = Q-W
from first law of thermodynamics
Q= W
total heat absorbed
Q= μRT_{0} (1-1/n)
= 2 x 8.314 x 300 x (1-1 / 2)
= 2494.25
= 2.5 KJ
(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^{(γ-1)/2} = constant = T_{0}V_{0}^{(γ-1)/2}
(T / T_{0}) = (V_{0} / V) ^{(γ-1)/2} = 1-(1 / η^{(γ-1)/2})
So, W = 2 Cv T_{0} (1 - T / T_{0})
=2 C_{V} T_{0} [1 - 1 / (η)^{(γ-1)/2}]
= (2RT_{0} / γ-1) [1 - 1 / (η)^{(γ-1)/2}]
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.
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
(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
a. Change in internal energy = C_{v}ΔT=RΔT/(y-1)
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/(1-y)
Similarly for PV^{n}=Constant
Work done =RΔT/(1-n)
Therefore work done in this case
W=RΔT/(1-2)
=-RΔT
c. Molar heat capacity is given by
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^{γ-1}=constant
T= T_{0}/3^{γ-1}
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
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/(γ-1)})[P_{0} + (W/A)]V_{0}/R
= [P_{0} + (W/A)][V_{0}/(γ-1)]
d Heat supplied
Q= ΔU +W
= [P_{0} + (W/A)][V_{0}/(γ-1)] + [P_{0} + (W/A)]V_{0}
=[P_{0} + (W/A)][V_{0}y/(γ-1)]
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^{γ-1}
b. For sample A
T_{0}V_{0}^{γ-1}=T_{A}(2V_{0})^{γ-1}
T_{A} = T_{0}/2^{γ-1}
for sample B
T_{B} = T_{0}
So T_{A} : T_{B} = 1 : 2^{γ-1}
c. Work done by the Sample A
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}
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
Gas |
No of moles |
Adiabatic coefficient |
A |
1 |
1.67 |
B |
2 |
1.4 |
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/y-1
=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})
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}^{γ-1}=T_{Y}V_{Y}^{γ-1}
T_{X}= T_{Y}(V_{Y}/V_{X})^{γ-1}
=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})/γ-1
=-5R/2* 609
W_{YZ}=nR(T_{Z}-T_{Y})
=909R
W_{ZA}= -nR(T_{Z}-T_{A})/γ-1
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}^{γ-1}=T_{A}V_{A}^{γ-1}
T_{A}=T_{Z}(V_{Z}/V_{A})^{γ-1}
=791K
So
W_{ZA} =5R/2 * 1027
Also W_{AX}=0 as volume constant
Then
W_{NET}=909R + 5R/2(1027-609)
=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}
(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.