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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
A quantity of air is kept in a container having walls which are slightly conducting. The initial temperature and volume are `27^0C` (equal to the temperature of the surrounding) and `800cm^(3)` respectively. Find the rise in the temperature of the gas is compressed to `200 cm^(3)` (a) in a short time (b) in a long time . Take gamma= `1.4`. |
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Answer» (a) When the gas is compressed in a short time, the process is adiabetic. Thus `T_2V_2^(gamma-1)=T_1V_1^(gamma-1)` `T_2=T_1((V_1)/(V_2))^(gamma-1)=(300)((800)/(200))^(1.5-1)=(300)(4)^(0.5)` `=600K` Rise in temperature `DeltaT=T_2-T_1=600-200=400K` (b) When the gas is compressed in a long time, the process is isothermal. `trangleT=0` |
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| 2. |
A gas is contained in a metallic cylinder fitted with a piston.the piston is suddenly moved in to compress the gas and is maintained at this position. As time passes the pressure of the gas in the cylinderA. increaseB. decreaseC. remains constantD. increase or decrease depending on the nature of the gas. |
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Answer» Correct Answer - B decrease |
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| 3. |
Five moles of an ideal monoatomic gas with an initial temperature of `127^@C` expand and in the process absorb 1200J of heat and do 2100J of work. What is the final temperature of the gas? |
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Answer» Correct Answer - A::C `DeltaU=Q-W=nC_VDeltaT=n(3/2R)(T_f-T_i)` `:.` `1200-2100=5(3/2xx8.31)(T_f-400)` `:.` `T_f=385K` `~~113^@C` |
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| 4. |
The ratio `(C_p)/(C_v)=gamma` for a gas. Its molecular weight is M. Its specific heat capacity at constant pressure isA. `(R)/(gamma-1)`B. `(gammaR)/(gamma-1)`C. `(gammaR)/(M(gamma-1))`D. `(gammaRM)/(gamma-1)` |
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Answer» Correct Answer - C `(C_P)/(C_V)=gamma`,`C_P-C_V=R` `C_V=(R)/(gamma-1)`,`C_P=C_V+R=(gammaR)/(gamma-1)` `s_P=(C_p)/(M)=(gammaR)/(M(gamma-1))` |
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| 5. |
A monoatomic gas of n-moles is heated temperature `T_1` to `T_2` under two different conditions (i) at constant volume and (ii) At constant pressure The change in internal energy of the gas isA. More for (i)B. More for (ii)C. Same in both casesD. Independent of number of moles |
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Answer» Correct Answer - C `DeltaU=nC_VDeltaT` (always) |
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| 6. |
One mole of an ideal gas goes from an initial state A to final state B via two processs : It first undergoes isothermal expansion from volume `V` to `3 V` and then its volume is reduced from `3 V` to `V` at constant pressure. The correct `P-V` diagram representing the two process in (figure)A. B. C. D. |
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Answer» Correct Answer - D On `P-V` diagram, isothermal process is represented by rectangular hyperbola. Isobaric process is straight line parallel to volume axis. |
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| 7. |
The first law of theromodynamics is a statement ofA. conservation of heatB. conservation of workC. conservation of momentumD. conservation of energy |
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Answer» Correct Answer - D conservation of energy |
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| 8. |
A monatomic gas expands at constant pressure on heating. The percentage of heat supplied that increases the internal energy of the gas and that is involed in the expansion isA. `75%,25%`B. `25%,75%`C. `60%,40%`D. `40%,60%` |
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Answer» Correct Answer - C `(DeltaU)/(DeltaQ)=(1)/(gamma)=(1)/((5)/(3))=0.6` i.e., `60%` `(DeltaW)/(DeltaQ)=100-60=40%` |
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| 9. |
Two moles of an ideal monoatomic gas are expanded according to the equation pT=constant form its initial state `(p_0, V_0)` to the final state due to which its pressure becomes half of the initial pressure. The change in internal energy is A. (a) `(3p_0V_0)/(4)`B. (b) `(3p_0V_0)/(2)`C. (c) `(9p_0V_0)/(2)`D. (d) `(5p_0V_0)/(2)` |
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Answer» Correct Answer - B Pressure becomes half. So, temperature is doubled. (as `Tprop1/p`) `DeltaU=nC_VDeltaT` `T_i=(p_0V_0)/(nR)` `T_f=(2p_0V_0)/(nR)` `:.` `Delta=T_f-T_i=(p_0V_0)/(nR)` `:.` `DeltaU=n(3/2R)((p_0V_0)/(nR))=(3p_0V_0)/(2)` |
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| 10. |
If the ratio of specific heat of a gas of constant pressure to that at constant volume is `gamma`, the change in internal energy of the mass of gas, when the volume changes from `V` to `2V` at constant pressure `p` isA. `(R)/(gamma-1)`B. `PV`C. `(PV)/((gamma-1))K`D. `(T-4)K` |
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Answer» Correct Answer - C `DeltaW=P(2V-V)=PV` `DeltaW=PDeltaV=nRDeltaT`( if P: constant) `nRDeltaT=PV` `DeltaU=nC_VDeltaT=(nRDeltaT)/(gamma-1)=(PV)/(gamma-1)` |
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| 11. |
An adiabatic vessel of total volume V is divided into two equal parts by a conducting separtor. The is fixed in this position. The part on the left contains one mole of an ideal gas `(U=1.5 nRT)` and the part on the right contains two moles of the same gas. initially, the pressure on each side is p. the system is left for sufficiant time so that a steady state is reached. find (a) the work done by the gas in the left part during the process, (b) the temperature on the two sides in the beginning, (c ) the final common temperature reached by the gases, (d) the heat given to the gas in the right part and (e) the increase in the internal energy of the gas in the left part. |
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Answer» Correct Answer - A::B::C::D (a) As the conducting wall is fixed the work done by the gas on the left part during the process is zero. (b) For left side, Let pressure `= P` `Volume = V` No of moles `= n (1 mole)` Let initial temperature `= T_(1) `(PV)/(2) = nRT_1` `rArr (PV)/(2) = (1)nRT` `rArr T_1 = (PV)/((2moles) R)` For Rightside No of moles` = (2moles). ` ` Let initial temperature` = T_2` (PV)/(2) = nRT_(2)` `T_2 = (PV)/(2moles)R)` Let final temperature = t Final pressure `= P` No of moles ` = 1mole+2mole` ` =3mole` `PV =nRT` ` ` ( PV)/(nR)` ` ` = (PV)/((3mole)xxR)` (d) For RHS, ` ` Delta Q =Delta U as Delta W = 0` ` Delta U = 1.5n_2R(T-T_2)` ` = 1.5xx2xxR(T-T_2)` ` 1.5xx2xx(4PV-3PV)/(4xx3 mole)` ` = (3xxPV)/(4xx3 moles) = (PV)/(4)` (e) As `dQ = -dU` ` rArr dU = -dQ = (-PV)/(4)` |
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| 12. |
Method 1 of Q Temperature of two moles of a monoatomic gas is increased by 300 K in the process `p prop V`. (a) Find molar heat capacity of the gas in the given process. (b) Find heat given to the gas in that. |
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Answer» Correct Answer - A::B (a) `p prop Vimplies pV^-1=` constant If we compare with `pV^x=` constant, then `x=-1` Now, `C=C_V+(R)/(1-x)` `C_V=3/2R` for a monoatomic gas `:.` `C=3/2R+(R)/(1-(-1))=2R` (b) `Q=nCDeltaT` Substituting the values, we get `Q=(2)(2R)(300)` `=1200R` |
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| 13. |
One mole of an ideal monoatomic gas at temperature `T_0` expands slowly according to the law `p/V` = constant. If the final temperature is `2T_0`, heat supplied to the gas isA. (a) `2RT_0`B. (b) `3/2RT_0`C. (c) `RT_0`D. (d) `1/2RT_0` |
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Answer» Correct Answer - A `pV^-1`=constant In the process, `pV^x=` constant `C=C_V+(R)/(1-x)` Here, `C=3/2R+(R)/(1+1)=2R` `Q=nCDeltaT=(1)(2R)(2T_0-T_0)` `=2RT_0` |
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| 14. |
P-V plots for two gases during adiabatic processes are shown in the figure. Plots 1 and 2 should corresponds respectively to A. `He` or `O_2`B. `O_2` and `He`C. `He` and `Ar`D. `O_2` and `N_2` |
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Answer» Correct Answer - B Higher value of `gamma`, curve is more steeper For monoatomic gas, `gamma_1=1.67`, for diatomic gas `gamma_2=1.4` `gamma_1gtgamma_2` i.e., Curve `2rarr` Monoatomic gas curve `1rarr` Diatoamic gas |
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| 15. |
p-V plots for two gases during adiabatic processes are shown in the figure. Plots 1 and 2 should correspond respectively to A. (a) `H_e` and `O_2`B. (b) `O_2` and `H_e`C. (c) `H_e` and `A_r`D. (d) `O_2` and `N_2` |
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Answer» Correct Answer - B In adiabatic process: slope of p-V graph, `(dp)/(dV)=-gammap/V` `slope prop gamma` (with negative sign) From the given graph, `(slope)_2 gt (slope)_1` `:.` `gamma_2 gt gamma_1` Therefore, 1 should correspond to `O_2(gamma=1.4)` and 2 should correspond to `H_e(gamma=1.67)`. Hence, the correct option is (b). |
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| 16. |
One mole of an ideal monatomic gas is taken round the cyclic process ABCA as shown in figure. Calculate (a) the work done by the gas. (b) the heat rejected by the gas in the path CA and the heat absorbed by the gas in the path AB, (c) the net heat absorbed by the gas in the path BC, (d) the maximum temperature attained by the gas during the cycle. |
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Answer» `n=1`,`C_V=(3R)/(2)`,`C_P=(5R)/(2)` `T_A=(P_0V_0)/(R)`,`T_B=(3P_0V_0)/(R)`,`T_C=(2P_0V_0)/(R)` (a) `DeltaW_(cyclic)=`area of ABC `=(1)/(2)(3P_0-P_0)(2V_0-V_0)=P_0V_0` (b) `DeltaQ_(CrarrA)=nC_P(T_A-T_C)=(5R)/(2)((P_0V_0)/(R)-(2P_0V_0)/(R))` `=-(5P_0V_0)/(2)` `DeltaQ_(ArarrB)=nC_V(T_B-T_A)=(3R)/(1)((3P_0V_0)/(R)-(P_0V_0)/(R))` `=3P_0V_0` (c) For cyclic process, `DeltaQ_(cyclic)=DeltaW_(cyclic)` `DeltaQ_(AB)+Delta_(BC)+DeltaQ_(CA)=DeltaW_(cyclic)` `3P_0V_0+DeltaQ_(BC)-(5P_0V_0)/(2)=P_0V_0` `DeltaQ_(BC)=(P_0V_0)/(2)` |
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| 17. |
A monoatomic ideal gas of two moles is taken through a cyclic process starting from A as shown in figure. The volume ratios are `(V_B)/(V_A)=2 and (V_D)/(V_A)=4`. If the temperature `T_A` at A is `27^@C`. Calculate, (a) the temperature of the gas at point B, (b) heat absorbed or released by the gas in each process, (c) the total work done by the gas during the complete cycle. Express your answer in terms of the gas constant R. |
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Answer» `n=2`, let `V_A=V_0`,`V_B=2V_0`,`V_D=4V_0=V_C` `T_A=27^@C=300K`,`C_V=(3R)/(2)`,`C_P=(5R)/(2)` (a) `ArarrB` isobaric process `(V_A)/(T_A)=(V_B)/(T_B)implies(V_0)/(300)=(2V_0)/(T_B)` `T_B=600K` (b) `ArarrB` (isobaric) `DeltaQ_(ArarrB)nC_PDeltaT=2.(5R)/(2)(T_B-T_A)` `=5R(600-300)=1500R` `BrarrC`(Isothermal) `DeltaQ_(BrarrC)=DeltaW_(BrarrC)=nRT_(B)ln((V_C)/(V_B))` `=2R(600)ln((4V_0)/(2V_0))` `=1200Rln2` (c) `DeltaQ_(cyclic)=DeltaQ_(ArarrB)+DeltaQ_(BrarrC)+DeltaQ_(CrarrD)+DeltaQ_(DrarrA)` `=600R=DeltaW_(cyclic)` |
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| 18. |
When a gas expands along AB, it does 500J of work and absorbs 250 J of heat. When the gas expands along AC, it does 700J of work and absorbs 300J of heat. (a) How much heat does the gas exchange along Bc? (b) When the gas makes the transition from C to A along CDA, 800J of work are done on it from C to D. How much heat does it exchange along CDA? |
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Answer» Correct Answer - A::B::D (a) `U_C-U_A` along two paths should be same. `:.` `(Q_(AB)+Q_(BC))-(W_(AB)+W_(BC))=Q_(AC)-W_(AC)` `:.` `(250+Q_(BC))-(500+0)=300-700` Solving, we get `Q_(BC)=-150J` (b) `Q_(CDA)=(W_(CD)+W_(DA))+(U_(A)_U_(C))` `=(-800+0)-(U_C-U_A)` `=-800-(300-700)` `=-400J` |
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| 19. |
Two moles of an ideal gas are undergone a cyclic process 1-2-3-1. If net heat exchange in the process is 300J, the work done by the gas in the process 2-3 is A. (a) `-500J`B. (b) `-5000J`C. (c) `-3000J`D. (d) None of these |
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Answer» Correct Answer - D 1-2 `TpropV` `:.` p=constant `Q_(n et)=W_(n et)=W_12+W_23+W_31` `:.` `W_23=Q_(n et)-W_12` (as `W_31=0`) `=-300-nRDeltaT` `=-300-2xx8.31xx300` `-5286J` |
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| 20. |
A steam engin intakes 100g of steam at `100^((0))C` per minute and cools it down to `20^((0))C`. Calculate the heat rejected by the steam engine per minute. Latent heart of vaporization of steam `=540 cal g^(-1)`. |
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Answer» Heat rejected during the condensation of steam in one minute `=(100 g)xx(540 cal g^(-1)=5.4xx10^(-4) cal`. Heat rejected during the cooling of water `=(100 g)xx(1 cal g^(-1)(0)C^(-1))(100(0)C-20(0)C)` `=8000 cal`. Thus, heat rejected by the engine per minute `=5.4xx10^(4) cal+0.8xx10^(4) cal` `=6.2xx10^(4)cal`. |
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| 21. |
For the thermodynamic cycle shown in figure find (a) net output work of the gas during the cycle, (b) net heat flow into the gas per cycle. |
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Answer» Correct Answer - A::B `Q_(n et)=W_(n et)=` area under the cycle `=(2xx10^5)(2.5xx10^-6)` `=0.5J` |
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| 22. |
A thermodynamic system is taken through the cycle `ABCD` as shown in the figure. Heat rejected by the gas during the cycle is A. `2PV`B. `4PV`C. `(1)/(2)PV`D. `PV` |
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Answer» Correct Answer - A `DeltaW_(cyclic)=-(2P-P)(3V-V)=-2PV` Heat rejected in cycle `ABCD=2PV` |
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| 23. |
100 moles of an ideal monatomic gas undergoes the thermodynamic process as shown in the figure `ArarrB:` isothermal expansion `BrarrC:` adiabatic expansion `CrarrD:` isobaric compression `DrarrA:` isochoric process The heat transfer along the process AB is `9xx10^4J`. The net work done by the gas during the cycle is [Take `R=8JK^-1mol^-1`]A. (a) `-0.5xx10^4J`B. (b) `+0.5xx10^4J`C. (c) `-5xx10^4J`D. (d) `+5xx10^4J` |
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Answer» Correct Answer - D `Q_(BC)=0` `Q_(CD)=nC_pDeltaT` `=n(5/2R)(T_D-T_C)` `=5/2(p_DV_D-p_CV_C)` `=5/2(10^5-2xx10^5)` `=-25xx10^4J` `Q_(DA)=nC_VDeltaT` `=n(3/2R)(T_A-T_D)` `=3/2(p_AV_A-p_DV_D)` `=3/2(2.4xx10^5-10^5)` `=21xx10^4J` Now, `W_(n et)=Q_(n et)` `=(9-25+21)xx10^4J=5xx10^4J` |
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| 24. |
A thermodynamic system is taken through the cycle abcda during the parts ab, bc, cd and da. (b) find the total heat rejected by the gas during the process. |
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Answer» (a) The work done during the part ab, `=int_(a)^(b)pdV=(100 kPa)int_(a)^(b) dV` `=(100 kPa)(300 cm^(3)-100 cm^(3)) `=20 J` The work done during bc is zero as the volume does not change. The work done during cd `=int_(c)^(d) pdV=(200 kPa)(100 cm^(3)-300 cm^(3)) ==-40 J)`. The work done during da is zero as the volume does not change. (b) The total work done by the system during the cycle abcda `DeltaW=20 J-40 J=-20 J`. The change in internal energy `DeltaU=0` as the initial state is the same as the final state. thus `DeltaQ=DeltaU+DeltaW=20 J`. so the system reject 20 J of heat during the cycle. |
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| 25. |
A thermodynamic system undergoes a cyclic process as shown in figure. The cycle consists of two closed loops, loop I and loop II. (a) Over one complete cycle, does the system do positive or negative work? (b) In each of loops I and II, is the net work done by the system positive or negative? (c) Over one complete cycle, does heat flow into or out of the system? (d) In each of loops I and II, does heat flow into or out of the system? |
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Answer» Correct Answer - A::B::C::D (a) `W_I=+ve` (as cycle is clockwise) `W_(II)=-ve` (as cycle is anti-clockwise) But, `|W_1|gt|W_(II)|` `:.` `W_(n et)=+ve` (c) `Q_(n et)=W_(n et)=+ve` (d) In loop-I, `Q_1=W_I=+ve` In loop-II, `Q_(II)=W_(II)=-ve` |
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| 26. |
A thermodynamic system undergoes a cyclic process as shown in figure. (a) over one complete cycle, does the system do positive or negative work. (b) over one complete cycle, does heat flow into or out of the system. (c) In each of the loops 1 and 2, does heat flow into or out of the system. |
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Answer» Correct Answer - A::B::C (a) `W_1=+ve`, as cycle is clockwise `W_2=`-ve as cycle is anti-clockwise Since, `|W_1|gt|W_2|` `:.` Net work done by the system is positive. (b) In cyclic process, `Q_(n et)=W_(n et)` Since, `W_(n et)` is positive. So `Q_(n et)` is also positive. So, heat flows into the system. (c) `W_1=Q_1=+ve`, so into the system. `W_2=Q_2=-ve`, so out of the system. |
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| 27. |
A thermodynamic system undergoes cyclic process `ABCDA` as shown in figure. The work done by the system is A. `P_0V_0`B. `2P_0V_0`C. `(P_0V_0)/(2)`D. zero |
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Answer» Correct Answer - D `DeltaW_(AODA)=(1)/(2)(2P_0-P_0)(2V_0-V_0)=(1)/(2)P_0V_0` `AODA` is clockwise cycle on `P-V` diagram, `DeltaW=+ve` `DeltaW_(COBC)=-(1)/(2)(3P_0-P_0)(2V_0-V_0)=(1)/(2)P_0V_0` `COBC` is anticlockwise cycle on `P-V` diagram, `DeltaW=-ve` `DeltaW_(ABCDA)=DeltaW_(AODA)+DeltaW_(COBC)=0` |
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| 28. |
Two moles of a monatomic ideal gas undergo a cyclic process ABCDA as shown in figure. BCD is a semicircle. Find the efficiency of the cycle. |
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Answer» Correct Answer - B Process AB is isochoric (V=constant). Hence, `DeltaW_(AB)=0` `DeltaW_(BCD)=p_0V_0+pi/2(p_0)(V_0/2)` `=(pi/4+1)p_0V_0` `DeltaW_(DA)=-1/2(p_0/2+p_0)(2V_0-V_0)` ` =-3/4p_0V_0` `DeltaU_(AB)=nC_VDeltaT=(2)(3/2R)(T_B-T_A)` `(n=2,C_V=3/2R)` `=3R((p_0V_0)/(2R)-(p_0V_0)/(4R))` `=3/4p_0V_0=DeltaQ_(AB)` `(T=(pV)/(nR))` `DeltaU_(BCD)=nC_VDeltaT=(2)(3/2R)(T_D-T_B)` `=(3R)((2p_0V_0)/(2R)-(p_0V_0)/(2R))=3/2p_0V_0` Hence, `DeltaQ_(BCD)=DeltaU_(BCD)+DeltaW_(BCD)` `(pi/4+5/2)p_0V_0` `DeltaU_(DA)=nC_VDeltaT` `=(2)(3/2R)(T_A-T_D)` `=(3R)((p_0V_0)/(4R)-(2p_0V_0)/(2R))` `=-9/4p_0V_0` `:.` `DeltaQ_(DA)=DeltaU_(DA)+DeltaW_(DA)` `=-9/4p_0V_0-3/4p_0V_0` `=-3p_0V_0` Net work done is, `W_(n et)=(pi/4+1-3/4)p_0V_0` `=1.04p_0V_0` and heat absorbed is `Q_(ab)=DeltaQ_(+ve)` `=(3/4+pi/4+5/2)p_0V_0=4.03p_0V_0` Hence, efficiency of the cycle is `eta=(W_(n et))/(Q_(ab))xx100` `=(1.04p_0V_0)/(4.03p_0V_0)xx100` `=25.8%` |
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| 29. |
An ideal monoatomic gas is taken round the cycle ABCDA as shown in the P-V diagram. The work done during the cycle is A. `(1)/(2)pV`B. `rhoV`C. `2rhoV`D. `4rhoV` |
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Answer» Correct Answer - d `W=(2V-V)(2p-p)=pV` |
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| 30. |
An ideal monatomic gas undergoes a process where its pressure is inversely proportional to its temperature. (a) Calculate the molar specific heat of the process. (b) Find the work done by two moles of gas if the temperature change from `T_1` to `T_2`. |
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Answer» Correct Answer - A::B::D (a) `pprop1/T` or `pT=const ant` `:.` `p(pV))=const ant` or `pV^(1//2)=const ant` In the process, `pV^x=const ant`, Molar heat capacity is `C=C_V+(R)/(1-x)` or `C=3/2R+(R)/(1-1/2)` `=3/2R+2R=7/2R` (b) `W=Q-DeltaU=nCDeltaT-nC_VDeltaT` `=n(C-C_V)DeltaT` `=2[7/2R-3/2R](T_2-T_1)` `=4R(T_2-T_1)` |
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| 31. |
A thermodynamical process is shown in the figure with `p_A=3xxp_(atm)`, `V_A=2xx10^-4m^3`, `p_B=8xxp_(atm)`, `V_C=5xx10^-4m^3`. In the process AB and BC, 600J and 200J heat are added to the system. Find the change in internal energy of the system in the process CA. `[1 p_(atm)=10^5N//m^2]` A. (a) `560J`B. (b) `-560J`C. (c) `-240J`D. (d) `+240J` |
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Answer» Correct Answer - B In a cyclic process, `Q_(n et)=W_(n et)` `:.` `Q_(AB)+Q_(BC)+Q_(CA)=` area under the graph `:.` `600+200+Q_(CA)=1/2(3xx10^-4)(5xx10^5)` `=75J` `:.` `Q_(CA)=-725J=W_(CA)+DeltaU_(CA)` `=-725=` (-Area under the graph)`+DeltaU_(CA)` `=-(1/2xx11xx10^5)(3xx10^-4)+DeltaU_(CA)` Solving we get, `DeltaU_(CA)=-560J` |
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| 32. |
The molar heat capacity for an ideal gas (i) Is zero for an adiabatic process (ii) Is infinite for an isothermal process (iii) depends only on the nature of the gas for a process in which either volume or pressure is constant (iv) Is equal to the product of the molecular weight and specific heat capacity for any processA. (i),(iii)B. (ii),(iii)C. (iii),(iv)D. all |
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Answer» Correct Answer - D (i) Adiabatic process `DeltaQ=0` `DeltaQ=nCDeltaT=0impliesC=0` (ii) Isothermal process, `DeltaT=0` `DeltaQ=nCDeltaTimpliesC=(DeltaQ)/(nDeltaT)=(DeltaQ)/(nxx0)=infty` (iii) P: Constant `C=C_P=(gammaR)/(gamma-1)` V: constant `C=C_V=(R)/(gamma-1)` `gamma:` depends on nature of gas (iv) `C=Ms` |
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| 33. |
Assertion: Isothermal and adiabatic, two processes are shown on p-V diagram. Process-1 is aidabatic and process-2 is isothermal. Reason: At a given point, slope of adiabatic process `=gammaxx`slope of isothermal process. 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 - A::B | |
| 34. |
calculate the change in internal energy of a gas kept in a rigid container when 100 J of heat is supplied to it. |
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Answer» Correct Answer - A Here, `Q=100J` `We know,`DeltaU=DeltaQ-DeltaW Here, since the container is rigid,` `(DeltaV)=0` ` Hence the `DeltaW=PDeltaV=0` `So,`DeltaU=DeltaQ=100 J. |
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| 35. |
A metal of mass 1 kg at constant atmospheric pressure and at initial temperature `20^@C` is given a heat of 20000J. Find the following (a) change in temperature, (b) work done and (c) change in internal energy. (Given, specific heat `=400J//kg-^@C`, cofficient of cubical expansion, `gamma=9xx10^-5//^@C`, density `rho=9000kg//m^3`, atmospheric pressure `=10^5N//m^2`) |
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Answer» Correct Answer - A (a) From `DeltaQ=msDeltaT` `Delta=(DeltaQ)/(ms)=(20000)/(1xx400)=50^@C` (b) `DeltaV=V_gammaDeltaT=(1/9000)(9xx10^-5)(50)` `=5xx10^-7m^3` `:.` `W=p_0*DeltaV=(10^5)(5xx10^-7)=0.05J` (c) `DeltaU=DeltaQ-W=(20000-0.05)J` `=19999.95J` |
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| 36. |
In a process, the pressure of an ideal gas is proportional to square of the volume of the gas. If the temperature of the gas increases in this process, then work done by this gasA. (a) is positiveB. (b) is negativeC. (c) is zeroD. (d) may be positive or negative |
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Answer» Correct Answer - A `ppropV^2` `:.` `T/VpropV` or `TpropV^3impliesVpropT^(1//3` T is increasing. So, V will also increase. Hence, work done will be positive. |
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| 37. |
The density `(rho)` versus pressure (p) graph of one mole of an ideal monoatomic gas undergoing a cyclic process is shown in figure. Molecular mass of gas is M. (a) Find work done in each process. (b) Find heat rejected by gas in one complete cycle. (c) Find the efficiency of the cycle. |
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Answer» Correct Answer - A::B::C For Process 1-2 `rho prop p` `:.` `1/Vprop p` `:.` Process is isothermal. `DeltaU_1=0` `Q_1=W_1=nRTIn((p_i)/(p_f))` `=(p_0M)/(rho_0)In(1/2)` (as `n=1` and `RT=(PM)/(rho)`) `=-(p_0M)/(rho_0)In(2)` For Process 2-3 `Q_2=Q_p=nC_pDeltaT` `=(1)(5/2R)(T_3-T_2)` `=5/2((p_3M)/(rho_3)-(p_2M)/(rho_2))` `=5/2((2p_0M)/(rho_0)-(2p_0M)/(2rho_0))` `=2.5(p_0M)/(rho_0)` `DeltaU_2=nC_VDeltaT` Substituting the values like above we get, `Delta_2=(1.5p_0M)/(rho_0)` `W_3=Q_2-DeltaU_2=(p_0M)/(rho_0)` For Process 3-1 Density is constant. Hence, volume is constant. `:.` `W_3=0` `:.` `Q_3=DeltaU_3=nC_VDeltaT` `=(1)(3/2R)(T_1-T_3)` `=3/2((p_0M)/(rho_0)-(2p_0M)/(rho_0))` `=-(1.5p_0M)/(rho_0)` `sumQ_(-ve)=|Q_1+Q_3|` `=(p_0M)/(rho_0)(3/2+In2)(+p_0M//rho_0)+((-p_0M)/(rho_0)In2)` (c) `eta=(W_(n et))/(sumQ_(+ve))=((+p_0M//rho_0)+((-p_0M)/(rho_0)In2))/((2.5p_0M//rho_0))` `=2/5(1-In2)` |
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| 38. |
50 cal of heat should be supplied to take a system from the state A to the state B through the path ACB as shown in Figure. The quantity of heat to be supplied to take it from A to B via ADB. |
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Answer» process ACB: `Delta_(ACB)=DeltaW_(AC)+DeltaW_(CB)` `=50xx10^3xx(400-200)xx10^-6+0` `=10J=(10)/(4.2)=2.4cal` `DeltaQ=50cal` `DeltaQ=DeltaU+DeltaW=U_B-U_A+DeltaW` `50=(U_B-U_A)+2.4` `U_B-U_A=47.6cal` Process ADB: `DeltaW_(ADB)=DeltaW_(AD)+DeltaW_(DB)` `=0+155xx10^3xx(400-200)xx10^-6` `=31J=(31)/(4.2)=7.4cal` `DeltaQ=DeltaU+DeltaW=U_B-U_A+7.4` `=47.6+7.4` `=55cal` |
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| 39. |
An ideal monoatomic gas undergoes a process in which its internal energy U and density `rho` vary as `Urho`= constant. The ratio of change in internal energy and the work done by the gas isA. (a) `3/2`B. (b) `2/3`C. (c) `1/3`D. (d) `3/5` |
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Answer» Correct Answer - A `Urho=` constant `:.` `T=(1/V)=const ant` or `TpropV` `:.` `p=const ant` `(DeltaU)/(W)=(DeltaU)/(Q-DeltaU)` `=(nC_VDeltaT)/(nC_pDeltaT-nC_VDeltaT)` `=(1)/(gamma-1)=(1)/(5/3-1)=3/2` |
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| 40. |
When an ideal monoatomic gas is heated at constant pressure find (a) Fraction of heat energy which increases the internal energy (b) Fraction of heat energy which is utlized as work. |
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Answer» `DeltaQ=DeltaU+DeltaW` (a) `(DeltaU)/(DeltaQ_P)=(nC_VDeltaT)/(nC_PDeltaT)=(1)/(gamma)=(1)/((5)/(3))=(3)/(5)` (b) `(DeltaW)/(DeltaQ_P)=1-(DeltaU)/(DeltaQ_P)=1-(3)/(5)=(2)/(5)` Or `(DeltaW)/(DeltaQ_P)=(PDeltaV)/(nC_PDeltaT)=(nRDeltaT)/(nC_PDeltaT)=(R)/((5R)/(2))=(2)/(5)` |
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| 41. |
When a system is taken through the process abc shown in figure, 80 J of heat is absorbed by the system and 30 J of work is done by it. If the system does 10 J of work during the process adc, how much heat flows into it during process? |
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Answer» Process abc: `DeltaQ=80J`,`DeltaW=30J`,`DeltaU=?` `DeltaQ=DeltaU+DeltaW` `80=DeltaU+30` `DeltaU=U_c-U_a=50J` process adc: `DeltaU=U_C-U_a=50J` `DeltaW=10J`,`DeltaQ=`? `DeltaQ=DeltaU+DeltaW+50+10=60J` |
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| 42. |
Method 2 of Q In a given process work done on a gas is 40 J and increases in its internal energy is 10J. Find heat given or taken to/from the gas in this process. |
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Answer» Correct Answer - C Given, `DeltaU=+10J` Work done on the gas is 40 J. Therefore, work done by the gas used in the equation, `Q=W+DeltaU` will be `-40J`. Now, putting the values in the equation, `Q=W+DeltaU` We have, `Q=-40+10` `=-30J` Here, negative sign indicates that heat is taken out from the gas. |
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| 43. |
Method 5 of W Heat taken from a gas in a process is 80 J and increase in internal energy of the gas is 20J. Find work done by the gas in the given process. |
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Answer» Correct Answer - A Heat is taken from the gas. Therefore, Q is negative. Or, `Q=-80J` Internal energy of the gas is increasing. Therefore, `DeltaU` is positive. Or `DeltaU=+20J` Using the first law eqution, `Q=W+DeltaU` or `W=Q-DeltaU=-80-20=-100J` Here, negative sign indicates that volume of the gas is decreasing and work is done on the gas. |
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| 44. |
Method 4 of W The temperature of n-moles of an ideal gas is increased from `T_0` to `2T_0` through a process `p=alpha/T`. Find work done in this process. |
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Answer» Correct Answer - B `pV=nRT` (ideal gas equation) …(i) and `p=alpha/T` …(ii) Dividing Eq. (i) by Eq. (ii), we get `V=(nRT^2)/(alpha` or `dV=(2nRT)/(alpha)dT` `:.` `W=int_(V_i)^(V_f)pdV=int_(T_0)^(2T_0)(alpha/T)((2nRT)/(alpha))dT` `=2nRT_0` |
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| 45. |
The internal energy of monoatomic ideal gas is 1.5 nRT. One mole of helium is kept in cylinder of cross section `8.5cm^2`. The cylinder is closed by a light frictionless piston. The gas is heated slowly in a process during which a total of 42 J heat is given to the gas. If the temperature rises through `2^@C` find the distance moved by the piston. Atmospheric pressure`=100kPa` |
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Answer» given `n=1`,`A=85xx10^-4m^2`,`DeltaQ=42J`,` DeltaT=2^@C` Atmospheric pressure `P_0=100Pa=10^5(N)/(m^2)`,`U=1.5nRT` The gas is heated slowly i.e., pressure remains constant `DeltaW=DeltaQ-DeltaU=42-1.5xx1xx8.3xx2=17.1J` `P_0DeltaV=P_0Ax=DeltaW` `10^5xx8.5xx10^-4x=17.1` `x=0.2m=20m` x distance moved by piston. |
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| 46. |
calculate the increase in internal energy of 1 kg of water at `100(0)C` when it is converted into steam at the same temperature and at 1atm (100 kPa). The density of water and steam are 1000 kg `m^(-3)` and 0.6 kg `m^(-3)` respectively. The latent heat of vaporization of water `=2.25xx10^(6) J kg^(1)`. |
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Answer» The volume of 1 kg of water `=(1)/(100)m^(3)` and of 1 kg of steam `=(1)/(0.6)m^(3)`. The increase in volume `=(1)/(0.6)m^(3)-(1)/(1000)m^(3)` `=(1.7-0.001)m^(3)~~1.7m^(3)`. The work done by the system is `pDeltaV` `=(100 kPa)(1.7 m^(3))` `=1.7xx10^(5) J`. the heat given to convert 1 kg of water into steam `2.25xx10^(6) J`. the change in internal energy is `DeltaU=DeltaQ-DeltaW` `=2.25xx10^(6) J-1.7xx10^(5) J` `=2.08xx10^(6) J`. |
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| 47. |
Calculate the increase in the internal energy of 10 g of water when it is heated from `0^(0)C to 100^(0)C` and converted into steam at 100 kPa. The density of steam `=0.6 kg m^(-3)` specific heat capacity of water `=4200 J kg^(-1 ^(0)C^(-3)` latent heat of vaporization of water `=2.25xx10 6 J kg^(-1)` |
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Answer» Correct Answer - A::B::D Mass` = 10kg = 0.01kg` `P = 10^5kpa` `Delta Q = QH_2 O 0^@-100^@+QH_(2) O-steam` `0.01xx4200xx100+0.01xx2.5xx10^3` ` = 4200+2500 = 29200` `Delta W = paV` `Delta V = ((0.01)/(06))-((0.01)/(1000)) = 0.01699` `DeltaW = pDeltaV = 0.01699xx10^5 = 1699J` ` = 27501 = 2.75xx10^4J |
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| 48. |
A gas is initinaly at a pressure of 100 kPa and its volume is `2.0 m^(3)`. Its pressure is kept constant and the volume is changed from `2.0 m^(3) to 2.5^(3)`. Its volume is now kept constant and the pressure is increased from 100 lPa to 200 kPa. The gas is brought back to its initial state, the pressure varying linearly with its volume. (a) whether the heat is supplied to or exterted from the gas in the complete cycle ? (b) how much heat was supplied or extracted ? |
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Answer» Correct Answer - A::B::C::D (a) `P_i = 100kPa, V_1 = 2m^3` `DeltaV_1 = 0.5m^3,Delta P_1 = 100kPa` `From the graph , we find that area under AC is greater then area under AB.So,we see that heat is extracted from the system (b)Amount of heat = area under Abc` ` = (1)/(2)xx(5)/(10)xx10^5 = 25000J |
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| 49. |
In a refrigerator, heat from inside at 277K is transferred to a room at 300K. How many joules of heat shall be delivered to the room for each joule of electrical energy consumed ideally? |
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Answer» Correct Answer - A::C Coefficient of performance of a refrigerator, `beta=Q_(2)/W=T_(2)/(T_(1)-T_(2))` `:. Q_(2)=WT_(2)/(T_(1)-T_(2))` But `W= "Energy consumed by the refrigerator "=1J, T_1=300K, T_2=277K` `:. Q_2=1xx(277)/(300-277)=(277)/(23)=12J` "Heat rejected by the refrigerator", `Q_1=W+Q_2=1+12` `=13J` |
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| 50. |
Assertion: In the process pT=constant, if temperature of gas is increased work done by the gas is positive. Reason: For the given process, `VpropT`.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. |
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Answer» Correct Answer - C `pT=` constant `:.` `(T/V)T=`constant or `VpropT^2` If temperature is increased, then volume will also increase. So, work done is positive. |
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