157 + Interview Questions in LAWS-OF-THERMODYNAMICS IN GENERAL AWARENESS Page 2 InterviewSolution

51.	The internal enegy of an ideal gas decreases nu the same amount as the work done by the system.A. The process must be adiabatic.B. The process must be isothermal.C. The process must be isobaric.D. The temperature must decrease.
Answer» Correct Answer - A::D The process must be adiabactive., The tempreture must decrease.

Discussion

52.	In a heat engine, the temperature of the source and sink are 500 K and 375 K. If the engine consumes `25xx10^5J` per cycle, find(a) the efficiency of the engine, (b) work done per cycle, and (c) heat rejected to the sink per cycle.
Answer» Correct Answer - A::B::C (a) `eta=1-(T_2)/(T_1)=1-375/500=0.25=25%` (b) `W=etaQ_1=0.25xx25xx10^5` `=6.25xx10^5J` (c) `Q_2=Q_1-W=(25-6.25)xx10^5` `=18.75xx10^5J`

Discussion

53.	the pressure of a gas change linearly with volume from 10kPa, 200 cc to 50 kPa, 50 cc. (a) calculate the work done by the gas, (b) what is the change in the internal energy of the gas ?
Answer» Correct Answer - A::B::D `P_10 kP_a` `=`10xx10^3P-a` `P_2=50xx10^3P_a` `V_1=200CC, V_2=50CC` `(i)work done on the gasltbrge`[((1)/(2))(10+50)xx10^3xx(50-200)xx10^-6]` `=`-`45 J` `(ii)`DeltaQ=0,` `So,lt `DeltaU=-DeltaW=45J`

Discussion

54.	A thermally insulaated, closed copper vessel contains water at `15^(0)C`. When the vessel is shaken vigorously for 15 minutes, the tempreture rises to `17^(0)C`. The mass of the vessel is 100 g and that of the water is 200 g. the specific heat capacities of copper and water are `420 Jkg^(-1) K^(-1)` and `4200 J kg^(-1)K^(-1)` respectively. negalect any thermal expansion. (a) how much heat is transferred to the liquid-vessel system ? (b)how much work has been done on this system? (c ) how much is the increase in internal energy if the system ?
Answer» Correct Answer - A::B::C::D `t_1=15^@C,t_2=17^@C,` `Deltat-t_2-t_1` `=`17^@C-15^@C=2^@C=275 K` `m_v=100g=01kg.,` `m_w=200g=02kg` ` Sp.heat capacity of` `c_u=420 J/kg-k` `Sp.heat capacity of water `=4200 J /kg-k (a) The heat transferred to the liquid vessel system is 0. The internal heat is shared in between the vessel and water. (b) Work done on the system = Heat product unit ` `rArr bw=100x10^-3x420x2+200` `x10^--3x4200x2` `=84+84x20=84x21` `=1764 J` ` (c)`dQ=0,dU=-dW=1764` ` since `dw =-ve work done in the system

Discussion

55.	In the previous question (i) work is done by the gas (ii) work is done on the gas (iii) heat is absorbed by the gas (iv) heat is given out by the gasA. (i),(iii)B. (ii),(iii)C. (iii),(iv)D. (i),(ii)
Answer» Correct Answer - C Clockwise cycle on `V-P` diagram `DeltaW_(cyclic)=-ve=DeltaQ_(cyclic)` Work is done on the gas, heat is given out by the gas

Discussion

56.	A gas undergoes a process such that `pprop1/T`. If the molar heat capacity for this process is `C=33.24J//mol-K`, find the degree of freedom of the molecules of the gas.
Answer» Correct Answer - D As `pprop1/T` or `pT=` constant …(i) We have for one mole of an ideal gas `pV=RT` …(ii) From Eqs. (i) and (ii), `p^2V=` constant or `pV^(1//2)=K (say) =p_iV_i^(1//2)=p_fV_f^(1//2)` …(iii) From first law of thermodynamics, `DeltaQ=DeltaU+DeltaW` or `CDeltaT=C_VDeltaT+DeltaW` or `C=C_V+(DeltaW)/(DeltaT)` ...(iv) Here, `DeltaW=intpdV=Kint_(V_i)^(V_f)V^(-1//2)dV` `=2K[V_f^(1//2)-V_i^(1//2)]=2[p_fV_f^(1//2)V_f^(1//2)-p_iV_i^(1//2)V_i^(1//2)]` `=2[p_fV_f-p_iV_i]=2R[T_f-T_i]` `=(RDeltaT)/(1//2)implies(DeltaW)/(DeltaT)=2R` Substituting in Eq. (iv), we have `C=C_V+2R=(R)/(gamma-1)+2R` Substituting the value, `33.24=R((1)/(gamma-1)+2)=8.31((1)/(gamma-1)+2)` Solving this we get `gamma=1.5` Now, `gamma=1+2/F` or degree of freedom `F=(2)/(gamma-1)=(2)/(1.5-1)=4`

Discussion

57.	0.2moles of an ideal gas, is taken around the cycle abc as shown in the figure. The path b-c is adiabatic process, a-b is isovolumic process and c-a is isobaric process. The temperature at a and b are `T_(A)=300K` and `T_(b)=500K` and pressure at a is 1 atmospere. find the volume at c. (Given, `gamma=(C_(p))/(C_(v))=(5)/(3),R=8.205xx10^(-2)L`/atm/mol-K) A. 6.9LB. 6.68LC. 5.52LD. 5.82L
Answer» Correct Answer - b From a-b, volume is constnat `(p_(s))/(T_(a))=(P_(b))/(T_(b))` `P_(b)=(500)/(300)xx1=(6)/(3)` atm For b-c, adiabatic process `(T_(c))/(T_(b))=((P_(c))/(P_(b)))^((gamma-1)/(gamma))=[(1)/(5//3)]^((5//3-1)/(5//3))=((3)/(5))^(2//5)` Also, `p_(b)V_(b)^(gamma)=p_(c)V_(c)^(gamma)` `V_(c)=(V_(b))((p_(b))/(p_(c)))^(1//gamma)` `=(4.928)((5//3)/(1))^(3//5)`=6.68litre

Discussion

58.	The state of an ideal gas is changed through an isothermal process at temperature `T_0` as shown in figure. The work done by the gas in going from state B to C is double the work done by gas in going from state A to B. If the pressure in the state B is `(p_0)/(2)`, then the pressure of the gas in state C is A. (a) `p_0/3`B. (b) `p_0/4`C. (c) `p_0/6`D. (d) `p_0/8`
Answer» Correct Answer - D `W=nRT_01n((p_i)/(p_f))` `W_(BC)=2W_(AB)` `:.` `nRT_0 In((p_C)/(p_0//2))=2nRT_0 In((p_0//2)/(p_0))` Solving this equation we get, `p_C=p_0/8`

Discussion

59.	(i) `DeltaW_(AB)=5P_0V_0` (ii) `DeltaW_(BC)=0` (iii) `DeltaW_(CA)=-2P_0V_0` (iv) `DeltaW_(ABCA)=3P_0V_0`A. (i),(iii)B. (ii),(iii)C. (iii),(iv)D. all
Answer» Correct Answer - D `DeltaW_(AB)=(1)/(2)(P_0+4P_0)(3V_0-V_0)=5P_0V_0` `DeltaW_(BC)=0` `DeltaW_(CA)=-P_0(3V_0-V_0)=-2P_0V_0` `DeltaW_(ABCA)=DeltaW_(AB)+DeltaW(BC)+DeltaW_(CA)=3P_0V_0`

Discussion

60.	Two moles of helium gas undergo a cyclic process as shown in Fig. Assuming the gas to be ideal, calculate the following quantities in this process (a) The net change in the heat energy (b) The net work done (c) The net change in internal energy
Answer» Correct Answer - A::B::C In a cyclic process, `DeltaU=0` `Q_(n et)=W_(n et)` `Q_(AB)=nC_pDeltaT` `=(2)(5/2R)(400-300)` `=500R` `Q_(BC)=nRT_(B)In(p_i/p_f)` `=(2)(R)(400)In(2/1)` `=800RIn(2)` `Q_(CD)=nC_pDeltaT` `=-500R` `Q_(DA)=nRT_DIn(p_i/p_f)` `=(2)(R)(300)In(1/2)` `=-600RIn(2)` `:.` `Q_(n et)=W_(n et)=(200R)In(2)` `=(200)(8.31)(0.693)` `~~1153J`

Discussion

61.	Find the molar specific heat of the process `p=a/T` for a monoatomic gas, a being constant.
Answer» Correct Answer - B We know that `dQ=dU+dW` Specific heat, `C=(dQ)/(dT)=(dU)/(dT)+(dW)/(dT)` …(ii) Since, `dU=C_(V)dT` …(ii) `C=C_(V)+(dW)/(dT)` `=C_(V)+(pdT)/(dT)` `p_V=RT` `:.` For the given process, `V=(RT)/p=(RT^2)/a` `(dV)/(dT)=(2RT)/a` `:.C=C_V+p((2RT)/(a))` `=C_V+2R` `=3/2R+2R=7/2R`

Discussion

62.	For a Carnot cycle (or engine) discussed in article 21.4, prove that efficiency of cycle is given by
Answer» Efficiency="net work done by gas"/"heat absorbed by gas" `=(\|W_1\|+\|W_2\|-\|W_3\|-\|W_4\|)/(\|Q_1\|)` …(i) Process 1 On this isothermal expansion process, the constant temperature is `T_1` so work done by the gas `W_1=nRT_1 In((V_b)/(V_a))` …(ii) Remember that `V_b gt V_a`, so this quantity is positive, as expected. (In process 1, the gas does work by lifting something) In isothermal process, `Q=W` `:.` `\|Q_1\|=\|W_1\|` ...(iii) Process 2 On this adiabatic expansion process, the temperature and volume are related through `TV^(gamma-1)=` constant `:.` `T_1V_b^(gamma-1)=T_2V_c^(gamma-1)` or `T_1/T_2=(V_c/V_b)^(gamma-1)` ...(iv) Work done by the gas in this adiabatic process is `W_2=(p_cV_c-p_bV_b)/(1-gamma)=(nRT_c-nRT_b)/(1-gamma)` `=((T_2-T_1)/(1-gamma))nR=((T_1-T_2)/(gamma-1))nR` ... (v) once again, as expected, this quantity is positive. Process 3 In the isothermal compression process, the work done by the gas is `W_3=nRT_(2) In(V_d)/(V_c)` ...(vi) Because `V_d gt V_c`, the work done by the gas is negative. The work done on the gas is `\|W_3\|=-nRT_(2) I n(V_d)/(V_c)=nRT_(2) In(V_c)/(V_d)` ...(vii) Furthermore, just as in process 1, `\|Q_3\|=\|W_3\|` ...(viii) Process 4 In the adiabatic compression process, the calculateions are exactly the same as they were in process 2 but course with different variables. Therefore, `T_2/T_1=(V_a/V_d)^(gamma-1)` ...(ix) and `W_4=(nR)/(gamma-1)(T_2-T_1)` ...(x) As expected, this quantity is negative and `\|W_4\|=-W_4=(nR)/(gamma-1)(T_1-T_2)` ...(xi) We can now calculate the efficiency. Efficiency `=(\|W_1\|+\|W_2\|-\|W_3\|-\|W_4\|)/(\|Q_1\|)` `=1+(\|W_2\|-\|W_3\|-\|W_4\|)/(\|Q_1\|)` (as `\|W_1\|=\|Q_1\|`) But our calculations show that `\|W_2\|=\|W_4\|`. Efficiency=`1-(\|W_3\|)/(\|Q_1\|)=1-(nRT_(2)ln(V_c//V_d))/(nRT_(1)ln(V_b//V_a))=1-(T_(2)ln(V_c//V_d))/(T_(1)ln(V_b//V_a))` ...(xii) We have seen that `(T_1)/(T_2)=(V_c/V_b)^(gamma-1)=(V_d/V_a)^(gamma-1)` or `V_c/V_b=V_d/V_a` or `V_c/V_d=V_b/V_a` Substituting in Eq. (xii), we get Efficiency `=1-T_(2)/T_(1)`

Discussion

63.	A Carnot engine works between 600K and 300K. The efficiency of the engine isA. (a) 50%B. (b) 70%C. (c) 20%D. (d) 80%
Answer» Correct Answer - A `eta=(1-(T_2)/(T_1))xx100` `T_2=300K` and `T_1=600K` `:.` `eta=50%`

Discussion

64.	Which of the following graphs correctly represents the variation of `beta=-(dV//dP)/V` with P for an ideal gas at constant temperature?A. B. C. D.
Answer» Correct Answer - A `PV=` constant `=nRT` `PdV+VdP=nRdT=0` `-(dV//dP)/(V)=(1)/(P)` `beta=(1)/(P)` `beta` v/s P will be rectangular hyperbola

Discussion

65.	In thermodynamic process, pressure of a fixed mass of a gas is changes in such a manner that the gas molecules gives out 20 J of heat and 10 J of work is done in the gas. If the initial internal energy of the gas was 40 J, then the final internal energy will beA. 30 JB. 20 JC. 60 JD. 40 J
Answer» Correct Answer - A `DeltaQ=-20J,DeltaW=-10J,U_i=40J,U_f=?` `DeltaQ=DeltaU+DeltaW=U_f-U_i+DeltaW` `-20=U_f-40-10` `U_f=30J`

Discussion

66.	Assertion: In adiabatic expansion, temperature of gas always decreases. Reason: In adiabatic process exchange of heat is zero.A. (a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. (b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. (c) If Assertion is true, but the Reason is false.D. (d) If Assertion is false but the Reason is true.
Answer» Correct Answer - B `Q=0`, `DeltaU=-W` In expansion, W is positive. So, `DeltaU` is negative. Hence, T also decreases.

Discussion

67.	A system can be taken from the initial state `p_(1),V_(1)` to the final state `p_(2,V_(2)` by two different methods, let `DeltaQ and DeltaW` represent the heat given to the system and the work done by the system. Which of the following must be the same in both the method?A. `DeltaQ`B. `DeltaW`C. `DeltaQ+DeltaW`D. `DeltaQ-DeltaW`.
Answer» Correct Answer - D `DeltaQ-DeltaW`.

Discussion

68.	In case of water from `0` to `4^@C` (i) Volume decreases and density of water is maximum at `4^@C` (ii) `DeltaW` will be negative, since volume decreases (iii) `C_p gt C_V` (iv) `C_P lt C_V`A. (i),(ii)B. (ii),(iii)C. (iii),(iv)D. (i),(ii),(iv)
Answer» Correct Answer - D `DeltaQ=DeltaU+DeltaW` since `DeltaW` is `-ve` i.e., `DeltaQ lt DeltaU` `nC_PDeltaT lt nC_VDeltaT` `C_P lt C_V`

Discussion

69.	When a system is taken from a state `i` to a state `f` in Figure, along the path `iaf`, we find `DeltaQ=50cal` and `DeltaW=20cal` while during the path `ibf`, `DeltaQ=36cal`. If `U_i=10cal`. (a) What is U? What is `W` along the path `ibf`? (c ) If `W=-13cal`, along the curved path `fi`, what is `DeltaQ` for this path? (d) If `U_b=22cal`, what is `DeltaQ` for the process `ib`? for `bf`?
Answer» `DeltaQ_(iaf)=50cal`,`DeltaW_(iaf)=20cal`,`DeltaQ_(ibf)=36cal` `U_i=10cal` (a) Path iaf: `DeltaU=U_f-U_k=U_f-10` `DeltaQ=DeltaU+DeltaW` `50=U_f-10+20impliesU_f=40cal` (b) Path ibf: `DeltaU=U_f-U_i=40-10=30cal` `DeltaQ=DeltaU+DeltaW` `36-30+DeltaWimpliesDeltaW=6cal` (c ) Path fi: `DeltaU=U_i-U_f=10-40=-30cal` `DeltaQ=DeltaU+DeltaW=-30-13=-43cal` Path ib: `DeltaU=U_b-U_i=22-10=12cal` `DeltaQ=DeltaU+DeltaW=12+6=18cal` `(DeltaW_(ib)=DeltaW_(ibf))` Path bf: `DeltaU=U_f-U_b=40-22=18cal` `DeltaQ=DeltaU+DeltaW=18+0=18cal`

Discussion

70.	A system can be taken from the initial state `p_(1),V_(1)` to the final state `p_(2,V_(2)` by two different methods, let `DeltaQ and DeltaW` represent the heat given to the system and the work done by the system. Which of the following must be the same in both the method?A. `DeltaQ`B. `DeltaW`C. `DeltaQ+DeltaW`D. `DeltaQ-DeltaW`
Answer» Correct Answer - D Since initial and final states are same for two processes `DeltaU_1=DeltaU_2=DeltaQ-DeltaW`

Discussion

71.	Consider two processes on a system as shown in figure. The volumes in the initial states are the same in the two processes and the volume in the final states are also the same.Let `DeltaW_(1)` and `DeltaW_(2)` be the work done by the system in the processes `A` and `B` respectively. A. `DeltaW_1 gt DeltaW_2`B. `DeltaW_1=DeltaW_2`C. `DeltaW_1 lt DeltaW_2`D. none
Answer» Correct Answer - C `DeltaW_1=P_ADeltaV,DeltaW_2=P_BDeltaV` `P_B gt P_AimpliesDeltaW_2 gt DeltaW_1`

Discussion

72.	In the figure given two processes `A and B` are shown by which a thermodynamic system goes from initial to final state F. if `DeltaQ_(A)` and `DeltaQ_(B)` are respectively the heats supplied to the systems then A. `DeltaQ_1 gt DeltaQ_2`B. `DeltaQ_1=DeltaQ_2`C. `DeltaQ_1 lt DeltaQ_2`D. `DeltaQ_1leDeltaQ_2`
Answer» Correct Answer - A `DeltaU_A=DeltaU_B,DeltaW_AgtDeltaW_B` (since area below the process `Agt` area below the process B) `DeltaQ_1=DeltaU_A+DeltaW_A,DeltaQ_2=DeltaU_B+DeltaW_B)` `DeltaQ_1gtDeltaQ_2`

Discussion

73.	A gas, for which `gamma` is `(4)/(3)` is heated at constant pressure. The percentage of heat supplied used for external work isA. `25%`B. `15%`C. `60%`D. `40%`
Answer» Correct Answer - A `(DeltaW)/(DeltaQ)=(nRDeltaT)/(nC_PDeltaT)=(R)/(C_P)=(R)/((gammaR)/((gamma-1)))` `=(gamma-1)/(gamma)=((4)/(3)-1)/((4)/(3))=(1)/(4)=0.25` i.e., `25%`

Discussion

74.	A refrigerator has to transfer an average of 263J of heat per second from temperature `-10^@C` to `25^@C`. Calculate the average power consumed, assuming no enegy losses in the process.
Answer» Correct Answer - C Here, `T_(1)=25+273=298K` `T_(2)=-10+273=263K` `Q_2=263J//s` Coefficient of performance is given by `beta=Q_(2)/(Q_(1)-Q_(2))=T_(2)/(T_1-T_2)` `:.` (Q_1)/(Q_2)=(T_1)/(T_2)` `:.` `Q_1=T_1/T_2xxQ_2=298/263xx263` `=298Js^-1` Average power consumed, `=298-263=35Js^-1=35W`

Discussion

75.	Find the work done by an ideal gas during a closed cycle `1rarr2rarr3rarr4rarr1` as shown in figure.
Answer» Divide the cylics process `1rarr2rarr3rarr4rarr1` in two processes, `1rarrOrarr2` and `3rarrOrarr4`. In two similar Deltas `12O` and `3O4` `(3P_0-P_0)/(5V_0-V_0)=(4P_0-3P_0)/(V_4-V_3)impliesV_4-V_3=2V_0` `DeltaW_(1rarrOrarr2)=(1)/(2)xx(3P_0-P_0)(5V_0-V_0)=4P_0V_0` `DeltaW_(3rarrOrarr4)=-(1)/(4)xx(4P_0-3P_0)(V_4-V_3)` `=-(P_0)/(2)xx2V_0=-P_0V_0` `DeltaW_(1rarr2rarr3rarr4rarr1)=4P_0V_0-P_0V_0` `=3P_0V_0`

Discussion

76.	Calculate the work done in the following processes, when gas expands from `V_0` to `2V_0` (a) `P=kV`, k is constant (b) `T=T_0+alphaV`,`T_0`,`alpha:` constant n: number of moles of gas
Answer» In case of variable quantities, work done can be calculated. `W=int_(V1)^(V2)PdV` (a) `W=int_(V0)^(2V0)kVdV=k\|(V^2)/(2)\|_(V0)^(2V0)` `=(k)/(2)[(2V_0)^2-V_0^2]` `=(3kV_(0)^(2))/(2)` (b) `T=T_0+alphaV` `(PV)/(nR)=T_0+alphaV` `P=(nRT_0)/(V)+alphanR` `W=int_(V0)^(2V0)PdV=int_(V0)^(2V0)((nRT_0)/(V)+alphanR)dV` `=nRT_0\|lnV\|_(v0)^(2v0)+alphanR\|V\|_(V0)^(2v0)` `nRT_0(ln2v_0-1nV_0)+alphanR(2V_0-V_0)` `nRT_0ln(2)+alphanRV_0`

Discussion

77.	consider the processes on a system shown in. during the processes,the work done by the system A. continuously increasesB. continuosly decreasesC. first increase then decreasesD. first decrease then increases then increase.
Answer» Correct Answer - A continuously increases

Discussion

78.	Method 3 of W Mass of a piston shown in Fig. is m and area of cross-section is A. Initially spring is in its natural length. Find work done by the gas.
Answer» Correct Answer - A::B In the given condition, work is done by the gas only against spring force `kx`. This force is a variable force. Hence, `W=int_0^x(kx)dx=1/2kx^2`

Discussion

79.	Method 2 of `DeltaU` Work done by a gas in a given process is `-20J`. Heat given to the gas is 60J. Find change in internal energy of the gas.
Answer» `DeltaU=Q-W` Substituting the values we have, `DeltaU=60-(-20)` `=80J` `DeltaU` is positive. Hence, internal erergy of the gas is increasing.

Discussion

80.	Make p-V equation for an adiabatic process.
Answer» Correct Answer - A::C In adiabatic process, `dQ=0` and `dW=-dU` `:.` `pdV=-C_VdT` (for `n=1`) `:.` `dT=-(pdV)/(C_V)` …(i) Also, for 1 mole of an ideal gas, `d(pV)=d(RT)` or `pdV+Vdp=RdT` or `dT=(pdV+Vdp)/(R)`…(ii) From Eqs. (i) and (ii), we get `C_VVdp+(C_V+R)pdV=0` or `C_VVdp+C_ppdV=0` Dividing this equaiton by PV, we are left with `C_V(dp)/(p)+C_p(dV)/(V)=0` or `(dp)/(p)+gamma(dV)/(V)=0` or `int(dp)/(p)+gamma int(dV)/(V)=0` or `In(p)+gamma In (V)=` constant We can write this in the form `pV^gamma=`constant

Discussion

81.	A cyclic process abcd is given for a monoatomic gas (`C_V=3/2R` and `C_p=5/2R`) as shown in figure. Find Q, W and `DeltaU` in each of the four processes separately. Also find the effieciency of cycle.
Answer» Correct Answer - A::C Process ab `V=` constant (`:.` Isochoric process) `W_(ab)=0` `:.` `Q_(ab)=DeltaU_(ab)=nC_VDeltaT` `=n(3/2R)(T_b-T_a)` `=3/2(nRT_b-nRT_a)` `=3/2(p_bV_b-p_aV_a)` `=3/2(2p_0V_0-p_0V_0)` `=1.5p_0V_0` Process bc `p=` constant (`:.` Isobaric process) `Q_(bc)=nC_pDeltaT` `=n(5/2R)(T_c-T_b)` `=5/2(nRT_c-nRT_b)` `=5/2(p_cV_c-p_bV_b)` `=5/2(4p_0V_0-2p_0V_0)` `=5p_0V_0` `DeltaU_(bc)=nC_VDeltaT` `=n(3/2R)(T_c-T_b)` `=3/2(nRT_c-nRT_b)` `=3/2(p_cV_c-p_bV_b)` `=3/2(4p_0V_0-2p_0V_0)` `=3p_0V_0` `W_(bc)=Q_(bc)-DeltaU_(bc)=2p_0V_0` Process cd Again an isochoric process. `:.` `W_(cd)=0` `Q_(cd)=DeltaU_(cd)=nC_VDeltaT` `=n(3/2R)(T_d-T_c)` `=3/2(nRT_d-nRT_c)` `=3/2(p_dv_d-p_cV_c)` `=3/2(2p_0V_0-4p_0V_0)` `=-3p_0V_0` Process da This is an isobaric process. `:.` `Q_(da)=nC_pDeltaT` `=n(5/2R)(T_a-T_d)` `=5/2(nRT_a-nRT_d)` `=5/2(p_aV_a-p_dV_d)` `=5/2(p_0V_0-2p_0V_0)` `=-2.5 p_0V_0` `DeltaU_(da)=nC_VDeltaT` `=n(3/2R)(T_a-T_d)` `=3/2(nRT_a-nRT_d)` `=3/2(p_aV_a-p_dV_d)` `=3/2(p_0V_0-2p_0V_0)` `=-1.5 p_0V_0` `W_(da)=Q_(da)-DeltaU_(da)` `=-p_0V_0` Efficiency of cycle In the complete cycle, `W_(n et)=W_(ab)+W_(bc)+W_(cd)+W_(da)` `=0+2p_0V_0+0-p_0V_0` `=p_0V_0`

Discussion

82.	Method 1 of `DeltaU` Temperature of two moles of a monoatomic gas is increased by 600 K in a given process. Find change in internal energy of the gas.
Answer» Correct Answer - A Using the equation, `DeltaU=nC_VDeltaT` for change in internal energy `C_V=3/2R` for monoatomic gas `DeltaU=(2)(3/2R)(600)` `=1800R`

Discussion

83.	The work done in the process AB isA. `8xx10^5J`B. `10xx10^5J`C. `7xx10^5J`D. `12xx10^5J`
Answer» Correct Answer - D On `P-V` diagram area below the process gives work done. `DeltaW_(AB)=[(1)/(2)(1+3)xx(3-1)+3xx(5-3)+(1)/(2)(1+3)(6-5)]xx10^5` `=(4+6+2)xx10^5=12xx10^5J`

Discussion

84.	Shows a process ABCA performed on an ideal gas. Find the net heat given to the system during the process.
Answer» As the process is cycle, the change in internal energy is zero. The heat given to the system is then equal to the work done byh it. The work done in part `AB is W_(1)=0` as the volume remains constant. The part BC represent an isothermal procces so that the work done by the gas during this part is `W_(2)=nRT_(2)In(V_(2)//V_(1))` during the part CA, `VpropT`. so, `V//T` is constant and hence, `p=(nRT)/V` is constant. The work done by the gas during the part CA is ` the work done by the gas the part CA is `W_(3)=p(V_(1)-V_(2))` `=nRT_(1)-nRT_(2)` `=-nR(T_(2)-T-(1). The net work done by the gas in the process ABCA is W=W_(1)+W_(2)+W_(3)=nR[T_(2)InV_(2)/V_(1)-(T_(2)-T_(1))]`. The same amount of heat is given to the gas.

Discussion

85.	One mole of an ideal monoatomic gas occupies a volume of `1.0xx10^-2m^3` at a pressure of `2.0xx10^5N//m^2`. (a) What is the temperature of tha gas? (b) The gas undergoes an adiabatic compression until its volume is decreased to `5.0xx10^-3m^3`. What is the new gas temperature? (c) How much work is done on the gas during the compression? (d) What is the change in the intenal energy of the gas?
Answer» Correct Answer - A::B::C::D (a) `T=(pV)/(nR)=((2xx10^5)(1.0xx10^-2))/((1)(8.31))` `=240.7K` (b) In adiabatic process, `TV^(gamma-1)=` constant `:.` `T_fV_f^(gamma-1)=T_iV_i^(gamma-1)` `:.` `T_f=T_i((V_i)/(V_f))^(gamma-1)` `=(240.7)[(1.0xx10^-2)/(5.0xx10^-3)]^(5//3-1)` `~~383K` (c) Work done on the gas `=DeltaU` =`nC_(V)Detla T` `=(1)(3/2R)(T_f-T_i)` `=1.5xx8.31(383-240.7)` `~~1770J` (d) `DeltaU~~1770J`

Discussion

86.	calculate the work done by a gas as it is taken from the state a to b, b to c and c to a as shown in
Answer» The work done by the gas in the process a to b is the area of adde. This is `W_(ab)=(120kPa)(250_(cc)` `=120xx10^(3)xx250xx10^(-6)J=30J`. In the process c to a the gas is compressed. The volume is decreased and the work done by the gas is negative.the magnitute is equel to the area of caed. this area is cab + baed `=(1)/(2)(20 kPa)(250 cc)+30J` `10J+30J=40J`. Thus, the work done in the process c to a is-40J.

Discussion

87.	An ideal gas at `27^(@)C` is compressed adiabatically to `8//27` of its original volume. If `gamma = 5//3`, then the rise in temperature isA. `450^@C`B. `375^@C`C. `225^@C`D. `405^@C`
Answer» Correct Answer - B `T_1=27+273=300K,V_1=V,V_2=(8)/(27)V,T_2=?` `T_1V_1^(gamma-1)=T_2V_2^(gamma-1)` `T_2=T_1((V_1)/(V_2))^(gamma-1)=(300)((V)/((8)/(27)V))^((5)/(3)-1)=(300)((27)/(8))^((2)/(3))` `=(300)[((3)/(2))^3]^((2)/(3))=(300)((3)/(2))^2` `=300xx(9)/(4)=675K` `DeltaT=T_2-T_1=675-300=375K=375^@C`

Discussion

88.	Consider the following two statements.(A) If heat is added to a system, its temperature must increase. (B) If positive work is done by a system in a thermodynamic process, its volume must increase.A. Both A and B are correctB. A is correct but B is wrongC. B is correct but A is wrongD. Both A and B are wrong
Answer» Correct Answer - C `DeltaQ=DeltaU=DeltaW` It may be that `DeltaQ` is utilized in doing work as in isothermal process, (l) is wrong. If `DeltaW=+ve`. Volume must increase

Discussion

89.	In a process on a system, the initial pressure and volume.A. The initial temperature must be equal to the final temperature.B. the initial internal energy must be equal to the final internal energy.C. the net heat given to the system in the process must be zero.D. the net work done by the system in the process must be zero.
Answer» Correct Answer - A::B The initial tempreture must be equal to the final tempreture. , the initial internal energy must be equal to the final internal energy.

Discussion

90.	The pressure p and volume V of an ideal gas both increase in a process.A. such a process is not possible.B. The work done by the system is positive.C. The tempreture of the system must increase.D. Heat supplied to the gas is equal to the change in internal energy.
Answer» Correct Answer - B::C The work done by the system is positive., The tempreture of the system must increase.

Discussion

91.	Assertion: The internal energy of an ideal gas does not change during an isothermal process. Reason: The decrease in volume of a gas is compensated by a corresponding increase in perssure, when its temp. is held constant.A. Both A and B are tue and R is the correct explanation of AB. Both A and R are true but R is not the correct explanation of AC. A is true but R is falseD. A is false but R is true
Answer» Correct Answer - A

Discussion

92.	A gas is contained in a cyclinder and expands according to the relation `rhoV^(13)`=constant. The initial pressure and initial volume of the gas are 30atm. And 30L respectively. If the final pressure is 15atm, then calculate the work done on the face of piston by the pressure force of the gas.A. `5xx10^(4)J`B. `4.36xx10^(4)J`C. `3xx10^(6)J`D. `4xx10^(4)J`
Answer» Correct Answer - b `W=int_(v_(1))^(2))` pdV and `pV^(1.3)=C.p=CV^(-13)` `W=int_(v_(1))^(v_(2))CV^(-1/3)dV=(-C)/(0.3)[V^(0.3)]_(v_(1))^(v_(2))` `C=P_(1)V_(1)^(1.3)=P_(2)V_(2)^(1.3)` `W=(p_(2)V_(2)-p_(1)V_(1))/(-0.3)` First volume, `V_(2)=V_(1)((p_(1))/(p_(2)))^(1//1.3)` `=30xx10^(-3)(2)^(0.77)=30xx10^(-3)xx1.71` `V_(2)=51.3xx10^(-3)m^(3)` `W=(15xx10^(5)xx51.3xx10^(-3)-30xx10^(5)xx30xx10^(-3))/(0.3)` `W=4.36xx10^(4)J`

Discussion

93.	If Q, E and W denote respectively the heat added, change in internal energy and the work done in a closed cycle process, thenA. `E=0`B. `Q=0`C. `W=0`D. `Q=W=0`
Answer» Correct Answer - A In cyclic process, `DeltaU=E=0`

Discussion

94.	When an ideal gas in a cylinder was compreswsed isothermally by a piston, the work done on the gas found to be `1.5xx10^(4)` cal. During this process aboutA. `3.6xx10^3 cal` of heat flowed out from the gasB. `3.6xx10^3 cal` of heat flowed into the gasC. `1.5xx10^4` cal of heat flowed into the gasD. `1.5xx10^4` cal of heat flowed out from the gas
Answer» Correct Answer - A In isothermal process, `DeltaQ=DeltaW` `DeltaQ=DeltaW=-1.5xx10^4J=(-15xx10^4)/(4.2)=-3.6xx10^3cal` `DeltaQ` is `-ve` i.e., heat is rejected by the gas.

Discussion

95.	Molar heat capacity is directly related toA. temperatureB. heat energyC. molecular structureD. mass
Answer» Correct Answer - c Since `C_(p)-C_(v)=R` `(C_(p))/(C_(v))=gamma` or `C_(p)=gammaC_(v)` `because C_(p)-C_(v)=R` `C_(v)(gamma-1)=R` `therefore C_(v)=(R)/((gamma-1))` Also, `C_(p)=(gammaR)/((gamma-1))` But R is universal constnat, the value of `gamma` depends upon molecular structure.

Discussion

96.	The molar heat capacity in a process of a diatomic gas if it does a work of `Q/4` when a heat of `Q` is supplied to it isA. `(2)/(5)R`B. `(5)/(2)R`C. `(10)/(3)R`D. `(6)/(7)R`
Answer» Correct Answer - c `dU=C_(v)dT=((5)/(2)R)dT` `rArr dT=(2(dU))/(5R)` From first law of thermodynamics, `dU=dQ=dW` or `dU=Q-(Q)/(4)=(3Q)/(4)` Now, molar heat capacity. `C=(dQ)/(dT)=(Q)/(2((dU)/(5R)))=(5QR)/(2((3Q)/(4)))=(10)/(3)R`

Discussion

97.	A monoatomic gas is supplied heat Q very slowly keeping the pressure constant. The work done by the gas isA. `(2)/(3)Q`B. `(3)/(5)Q`C. `(2)/(5)Q`D. `(1)/(5)Q`
Answer» Correct Answer - C `DeltaQ=DeltaU+DeltaW` `=nC_VDeltaT+nRDeltaT` `=n.(3R)/(2).DeltaT+nRDeltaT` `Q=(5nRDeltaT)/(2)` `nRDeltaT=(2Q)/(5)=DeltaW`

Discussion

98.	An ideal gas of adiabatic exponent `gamma` is expanded so that the amount of heat transferred to the gas is equal to the decrease of its internal energy. Then, the equation of the process in terms of the variables T and V isA. (a) `TV^(((gamma-1))/(2))=C`B. (b) `TV^(((gamma-2))/(2))=C`C. (c) `TV^(((gamma-1))/(4))=C`D. (d) `TV^(((gamma-2))/(4))=C`
Answer» Correct Answer - A `dQ=-dU` or `dU+dW=-dU` or `2dU+dW=0` `:.` `2[nC_VdT]+pdV=0` or `2[n((R)/(gamma-1))dT]+pdV=0` or `(2nRdT)/(gamma-1)+((nRT)/(V))dV=0` or `((2)/(gamma-1))(dT)/(T)+(dV)/(V)=0` Integrating we get, `((2)/(gamma-1))In(T)+In(V)=In(C)` Solving we get `TV^((gamma-1)/(2))=const ant`

Discussion

99.	consider two processes on a system on a system as shown in the volumes in the initial states are the same in the two processes and the volume in the final states are also the same.Let `DeltaW_(1) and DeltaW_(2)` be the work done by the sustem in the processes A and B respectively. A. `DeltaW_(1)gtDelta_(2)`B. `DeltaW_(1)=DeltaW_(2)`C. `DeltaW_(1)ltDeltaW_(2)`D. nothing can be said about the relation between `DeltaW_(1) and DeltaW_(2)`.
Answer» Correct Answer - C `DeltaW_(1)ltDeltaW_(2)`

Discussion

100.	A 1.0 kg bar of copper is heated at atmospheric pressure `(1.01xx10^5N//m^2)`. If its temperature increases from `20^@C` to `20^@C`, calculate the change in its internal energy. `alpha=7.0xx10^-6//^@C`, `rho=8.92xx10^3kg//m^3` and `c=387J//kg-^@C`.
Answer» Correct Answer - A::B `V=(m)/(rho)=(1.0)/(8.92xx10^3)` `=1.12xx10^-4m^3` `gamma=3alpha=2.1xx10^-5per^@C` `DeltaV=VgammaDeltatheta` `=(1.12xx10^-4)(2.1xx10^-5)(30)` `=7.056xx10^-8m^3` `W=pDeltaV` `=(1.01xx10^5)(7.056xx10^-8)` `=7.13xx10^-3J` `Q=mcDeltaQ` =(1)(387)(30)` `Delta U=Q-W` `=11609.99287J`

Discussion

Explore topic-wise InterviewSolutions in .

The internal enegy of an ideal gas decreases nu the same amount as the work done by the system.A. The process must be adiabatic.B. The process must be isothermal.C. The process must be isobaric.D. The temperature must decrease.

In a heat engine, the temperature of the source and sink are 500 K and 375 K. If the engine consumes `25xx10^5J` per cycle, find(a) the efficiency of the engine, (b) work done per cycle, and (c) heat rejected to the sink per cycle.

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For a Carnot cycle (or engine) discussed in article 21.4, prove that efficiency of cycle is given by

A Carnot engine works between 600K and 300K. The efficiency of the engine isA. (a) 50%B. (b) 70%C. (c) 20%D. (d) 80%

Which of the following graphs correctly represents the variation of `beta=-(dV//dP)/V` with P for an ideal gas at constant temperature?A. B. C. D.

In thermodynamic process, pressure of a fixed mass of a gas is changes in such a manner that the gas molecules gives out 20 J of heat and 10 J of work is done in the gas. If the initial internal energy of the gas was 40 J, then the final internal energy will beA. 30 JB. 20 JC. 60 JD. 40 J

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A gas, for which `gamma` is `(4)/(3)` is heated at constant pressure. The percentage of heat supplied used for external work isA. `25%`B. `15%`C. `60%`D. `40%`

A refrigerator has to transfer an average of 263J of heat per second from temperature `-10^@C` to `25^@C`. Calculate the average power consumed, assuming no enegy losses in the process.

Find the work done by an ideal gas during a closed cycle `1rarr2rarr3rarr4rarr1` as shown in figure.

Calculate the work done in the following processes, when gas expands from `V_0` to `2V_0` (a) `P=kV`, k is constant (b) `T=T_0+alphaV`,`T_0`,`alpha:` constant n: number of moles of gas

consider the processes on a system shown in. during the processes,the work done by the system A. continuously increasesB. continuosly decreasesC. first increase then decreasesD. first decrease then increases then increase.

Method 3 of W Mass of a piston shown in Fig. is m and area of cross-section is A. Initially spring is in its natural length. Find work done by the gas.

Method 2 of `DeltaU` Work done by a gas in a given process is `-20J`. Heat given to the gas is 60J. Find change in internal energy of the gas.

Make p-V equation for an adiabatic process.

A cyclic process abcd is given for a monoatomic gas (`C_V=3/2R` and `C_p=5/2R`) as shown in figure. Find Q, W and `DeltaU` in each of the four processes separately. Also find the effieciency of cycle.

Method 1 of `DeltaU` Temperature of two moles of a monoatomic gas is increased by 600 K in a given process. Find change in internal energy of the gas.

The work done in the process AB isA. `8xx10^5J`B. `10xx10^5J`C. `7xx10^5J`D. `12xx10^5J`

Shows a process ABCA performed on an ideal gas. Find the net heat given to the system during the process.

calculate the work done by a gas as it is taken from the state a to b, b to c and c to a as shown in

An ideal gas at `27^(@)C` is compressed adiabatically to `8//27` of its original volume. If `gamma = 5//3`, then the rise in temperature isA. `450^@C`B. `375^@C`C. `225^@C`D. `405^@C`

The pressure p and volume V of an ideal gas both increase in a process.A. such a process is not possible.B. The work done by the system is positive.C. The tempreture of the system must increase.D. Heat supplied to the gas is equal to the change in internal energy.

If Q, E and W denote respectively the heat added, change in internal energy and the work done in a closed cycle process, thenA. `E=0`B. `Q=0`C. `W=0`D. `Q=W=0`

Molar heat capacity is directly related toA. temperatureB. heat energyC. molecular structureD. mass

The molar heat capacity in a process of a diatomic gas if it does a work of `Q/4` when a heat of `Q` is supplied to it isA. `(2)/(5)R`B. `(5)/(2)R`C. `(10)/(3)R`D. `(6)/(7)R`

A monoatomic gas is supplied heat Q very slowly keeping the pressure constant. The work done by the gas isA. `(2)/(3)Q`B. `(3)/(5)Q`C. `(2)/(5)Q`D. `(1)/(5)Q`

A 1.0 kg bar of copper is heated at atmospheric pressure `(1.01xx10^5N//m^2)`. If its temperature increases from `20^@C` to `20^@C`, calculate the change in its internal energy. `alpha=7.0xx10^-6//^@C`, `rho=8.92xx10^3kg//m^3` and `c=387J//kg-^@C`.