157 + Interview Questions in LAWS-OF-THERMODYNAMICS IN GENERAL AWARENESS Page 3 InterviewSolution

101.	An ideal gas is initially at temperature T and volume V. Its volume is increased by `DeltaV` due to an increase in temperature `DeltaT,` pressure remaining constant. The quantity `delta=(DeltaV)/(VDeltaT)` varies with temperature asA. B. C. D.
Answer» Correct Answer - C `(V)/(T)=(nR)/(P)=`constant `(V)/(T)=(V+DeltaV)/(T+DeltaT)` `VT+VDeltaT=VT+DeltaVT` `VDeltaT=DeltaVT` `(DeltaV)/(VDeltaT)=(1)/(T)` `delta=(1)/(T)` `delta` v/s T will be rectangular hyperbola

Discussion

102.	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)`
Answer» Correct Answer - A::B::D `V_i=m/rho_i=0.01/1000=10^-5m^3` `V_f=m/rho_f=0.01/0.6=0.0167m^3` `DeltaU=Q-W` `=msDeltatheta+mL-p_0DeltaV` `=(0.01xx4200xx100)+(0.01)(2.5xx10)^6-10^5(0.0167-10^-5)` `=27530J`

Discussion

103.	The internal energy of a monatomic ideal gas is 1.5 nRT. One mole of helium is kept in a cylinder of cross section `8.5 cm^(2)`. The cylinder is closed by a light frictionless piston. The gas is heated slowly in a process during which a total of 42J heat is given to the gas. if the temperature rise through `2^((0))C`, find the distance moved by the piston. atmoshphere pressure `=100 kPa.
Answer» The change in internal energy of the gas is `DeltaU=1.5 nR(DeltaT)` `=1.5(1 mol)(8.3 JK^(-1))(2 K)` `24.9 J`. time heat given to the gas `=42J`. the work done by the gas is `DeltaW=DeltaQ-DeltaU` `=42 J-24.9 J=17.1 J`. if the distance moved by the piston is x, the work done is `DeltaW=(100 kPa)(8.5 cm^(2))x`. Thus, `(10^(5)Nm^(-2)) (8.5xx10^(-4)m^(-2))x=17.1 J` or, `x=0.2m=20 cm`.

Discussion

104.	As shown in figure three paths through which a gas can be taken from the state A to the state B. Calculate the work done by the gas in each of the three paths.
Answer» `DeltaW_(ACB)=DeltaW_(AC)+DeltaW_(CB)` `=0+30xx10^3xx(25-10)xx10^-6` `=0.4J` `DeltaW_(AB)=(1)/(2)(10+30)xx10^3xx(25-10)xx10^-6` `=0.30J` `DeltaW_(ADB)=DeltaW_(AD)+DeltaW(DB)` `=10xx10^3xx(25-10)xx10^-6+0` `=0.15J`

Discussion

105.	For an ideal gas the molar heat capacity varies as `C=C_V+3aT^2`. Find the equation of the process in the variables (T,V) where a is a constant.
Answer» Correct Answer - A::B::C `dQ=dU+dW` `CdT=C_VdT+pdV` `(C_V+3aT^2)dT=C_VdT+pdV` `:.` `3aT^2dT=pdV=((RT)/(V))dV` `:.` `((3a)/(R))TdT=(dV)/(V)` Integrating, we get `((3aT^2)/(2R))=InV-InC` `V=Ce^((3aT^2)/(2R))` or `Ve^(-(3aT^2)/(2R))`=const ant`

Discussion

106.	The molar heat capacity for an ideal gas cannotA. cannot be negativeB. must be equal to either `C_v` or `C_p`C. must lie in the range `C_vleCleC-p`D. may have any value between `-infty` and `+infty`
Answer» Correct Answer - D

Discussion

107.	In the cyclic process shown in the `V-P` diagram the magnitude of the work is done isA. `pi((P_1-P_2)/(2))^2`B. `pi((V_1-V_2)/(2))^2`C. `(pi)/(4)(P_2-P_1)(V_2-V_1)`D. `pi(P_2V_2-P_1V_1)`
Answer» Correct Answer - C `DeltaW_(cyclic)=-pi((P_2-P_1)/(2))((V_2-V_1)/(2))` `=-(pi)/(4)(P_2-P_1)(V_2-V_1)`

Discussion

108.	Let `(C_v) and (C_p)` denote the molar heat capacities of an ideal gas at constant volume and constant pressure respectively . Which of the following is a universal constant?A. `(C_p)/(C_v)`B. `C_pC-v`C. `C_p-C_v`D. `C_p+C_v`
Answer» Correct Answer - C

Discussion

109.	Each molecule of a gas has `f` degrees of freedom. The ratio `gamma` for the gas isA. `1+(f)/(2)`B. `1+(1)/(f)`C. `1+(2)/(f)`D. `1+((f-1))/(3)`
Answer» Correct Answer - C

Discussion

110.	A thermally insulated container is divided into two parts by a screen. In one part the pressure and temperature are `P` and `T` for an ideal gas filled. In the second part it is vacuum. If now a small hole is created in the screen, then the temperature of the gas willA. decreaseB. increaseC. remains sameD. none of the above
Answer» Correct Answer - C

Discussion

111.	A cornot engine in 1000 kilocal of heat from a reservoir at `627^@C` and exhausts it to sink at `27^@C`. What is its efficiency? How much work does it perform?
Answer» Given `T_1=627^@C=627+273=900K` `T_2=27^@C=27+273=300K` `Q_1=3000`kcal (a) `eta=1-(T_2)/(T_1)=1-(300)/(900)=(2)/(3)` i.e., `(200)/(3)=66.6%` (b) `(Q_2)/(Q_1)=(T_2)/(T_1)implies(Q_2)/(3000)=(300)/(900)` `Q_2=1000kcal` `W=Q_1-Q_2=3000-1000=2000kcal` `=2000xx10^3xx4.2=8.4xx10^6J`

Discussion

112.	A Carnot engine takes `3xx10^6cal`. of heat from a reservoir at `62^@C`, and gives it to a sink at `27^@C`. The work done by the engine is
Answer» Correct Answer - A::B::C `T_1=627+273=900K` `T_2=27+273=300K` `Q_1=3xx10^6cal` `eta=(1-(T_2)/(T_1))=(1-Q_2/Q_1)` `:.` `Q_2/Q_1=T_2/T_1` `Q_2=T_2/T_1xxQ_1` `(3xx10^6xx300)/(900)=10^6cal`

Discussion

113.	A rigid container of negligible heat capacity contains one mole of an ideal gas. The temperatures of the gas increases by `1^@C` if `3.0` cal of heat is added to it. The gas may beA. (i),(ii)B. (ii),(iii)C. (iii),(iv)D. (i),(iv)
Answer» Correct Answer - A Rigid container i.e., volume is constant `n=1`,`DeltaQ=3cal`,`DeltaT=1^@C` `(DeltaQ)_V=nC_VDeltaT` `3=1xxC_Vxx1` `C_V=3(cal)/(mol e^@C)` `C_P=C_V+R=3+2=5(cal)/(mol e^@C)` `gamma=(C_P)/(C_V)=(5)/(3)`, i.e., monoatomic gas.

Discussion

114.	A perfect gas goes from a state A to another state B by absorbing 8 × 105 J of heat and doing 6.5 × 105 J of external work. It is now transferred between the same two states in another process in which it absorbs 105 J of heat. In the second processA. Work done on the gas is `0.5xx10^5J`B. Work done by gas is `0.5xx10^5J`C. Work done on gas is `10^5J`D. Work done by gas is `10^5J`
Answer» Correct Answer - A Process 1: `DeltaQ=DeltaU+DeltaW` `8xx10^5=DeltaU+6.5xx10^5` `DeltaU=1.5xx10^5J` Process 2: `DeltaQ=DeltaU+DeltaW` `10^5=1.5xx10^5+DeltaWimpliesDeltaW=-0.5xx10^5J`

Discussion

115.	An ideal gas goes from the state `i` to the state `f` as shown in the figure.The work done by the gas during the process A. is positiveB. is negativeC. is zeroD. none
Answer» Correct Answer - C An incline line passing through origin on `P-T` diagram is isochoric process (V: constant) `DeltaW=0`

Discussion

116.	In an adiabatic change, the pressure p and temperature T of a diatomic gas are related by the relation `ppropT^alpha`, where `alpha` equalsA. (a) 1.67B. (b) 0.4C. (c) 0.6D. (d) 3.5
Answer» Correct Answer - D `pT^((gamma)/(1-gamma))=constant` or `ppropT^((gamma)/(gamma-1))` `:.` `alpha=(gamma)/(gamma-1)=(1.4)/(1.4-1)=3.5`

Discussion

117.	The adiabatic elasticity of hydrogen gas `(gamma=1.4)` at `NTP`A. `1xx10^5(N)/(m^2)`B. `1xx10^8(N)/(m^2)`C. `1.4(N)/(m^2)`D. `1.4xx10^5(N)/(m^2)`
Answer» Correct Answer - D `K_(adi)=gammaP=(1.4)(10^5)=1.4xx10^5(N)/(m^2)`

Discussion

118.	In an adiabatic change, the pressure p and temperature T of a diatomic gas are related by the relation `ppropT^alpha`, where `alpha` equalsA. `(5)/(3)`B. `(2)/(5)`C. `(3)/(5)`D. `(5)/(2)`
Answer» Correct Answer - D For adiabetic charge `(T^(gamma))/(P^(gamma-1))=`constant `T^(gamma)propP^(gamma-1)` `T^((gamma)/(gamma-1))proP` `PpropT^((gamma)/(gamma-1))` `PpropT^C` `c=(gamma)/(gamma-1)=(5//3)/((5)/(3)-1)=(5)/(2)`

Discussion

119.	The work of 146 kJ is performed in order to compress one kilo mole of a gas adiabatically and in this process the temperature of the gas increases by `7^@C`. The gas is `(R=8.3ml^-1Jmol^-1K^-1)`A. TriatomicB. A mixture of monoatomic and idatomicC. MonoatomicD. Diatomic
Answer» Correct Answer - D `n=10^3,DeltaW=-146xx10^3J,DeltaT=T_2-T_1=7^@C` `DeltaW=(nR(T_1-T_2))/(gamma-1)=-(nR(T_2-T_1))/(gamma-1)` `-146xx10^3=-(10^3xx8.3xx7)/(gamma-1)` `gamma-1=0.4` `gamma=1.4`, diatomic gas

Discussion

120.	A gas undergoes a process in which its pressure `P` and volume `V` are related as `VP^(n) =` constant. The bulk modulus of the gas in the process is:A. `np`B. `p^((1)/(n))`C. `(p)/(n)`D. `p^n`
Answer» Correct Answer - C `Vp^n=`constant `Vnp^(n-1)dP+p^ndV=0` `(dp)/(dV)=-(p)/(nV)` Bulk modulus`=-(dp)/((dV)/(V))=-V(dp)/(dV)` `=(-V)(-(P)/(nV))` `=(P)/(n)`

Discussion

121.	During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio `C_P//C_V` for the gas isA. 2B. 1.5C. `(5)/(3)`D. `(4)/(3)`
Answer» Correct Answer - B `PpropT^3` `(gamma)/(gamma-1)=3` (see solution of previous problem) `gamma=3gamma-3impliesgamma=(3)/(2)=1.5`

Discussion

122.	A gas follows a process `TV^(n-1)=const ant`, where `T=` absolute temperature of the gas and `V=` volume of the gas. The bulk modulus of the gas in the process is given byA. (a) `(n-1)p`B. (b) `p//(n-1)`C. (c) `np`D. (d) `p//n`
Answer» Correct Answer - C `TV^(n-1)=` constant `:.` `(pV)V^(n-1)=` constant (as `TproppV`) `:.` `pV^n=` constant Bulk modulus in the process `pV^gamma=` constant is `gammap`. Hence, in the given process Bulk modulus is np.

Discussion

123.	An ideal gas expands while the pressure is kept constant. During the process, does heat flow into the gas or out of the gas? Justify you answer.
Answer» Correct Answer - A `p=` constant `:.` `TpropV` As the gas expands V increases. So, T also increase. Hence, `DeltaT` is positive. Therefore, in the expression. `Q=nC_pDeltaT` Q is positive.

Discussion

124.	A sample of ideal gas is expanded to twice its original volume of `1.00m^3` in a quasi-static process for which `p=alphaV^2`, with `alpha=5.00 atm//m^6`, as shown in Fig. How much work is done by the expanding gas?
Answer» Correct Answer - A `W=int_(V_i)^(V_f)pdV=int_1^2(alphaV^2)dV` `=int_1^2(5xx1.01xx10^5)V^2dV` Solving we get, `W=1.18xx10^6J` `=1.18MJ`

Discussion

125.	Show how internal energy U varies with T in isochoric, isobaric and adiabatic process?
Answer» Correct Answer - A::B::C::D For all process, `DeltaU=nC_(V)DeltaT` `=(nRDeltaT)/(gamma-1)` (as `C_V=(R)/(gamma-1)`)

Discussion

126.	Refer to figure in previous question, `DeltaU_(1)` and `DeltaU_(2)` be the changes in internal energy of the system in the processes `A` and `B` thenA. `DeltaU_1 gt DeltaU_2`B. `DeltaU_1=DeltaU_2`C. `DeltaU_1 lt DeltaU_2`D. `DeltaU_1neDeltaU_2`
Answer» Correct Answer - B `DeltaU_1=DeltaU_2` Internal energy is state function.

Discussion

127.	Refer to let `DeltaU_(1)` and `DeltaU_(2)` be the changes in internal energy of the system in the processes A and B. thenA. `DeltaU_(1)gtDeltaU_(2)`B. `DeltaU_(1)=DeltaU_(2)`C. `DeltaU_(1)ltDeltaU_(2)`D. `DeltaU_(1)!=DeltaU_(2)`
Answer» Correct Answer - B `DeltaU_(1)=DeltaU_(2)`

Discussion

128.	Method 1 of W By integration, make expressions of work done by gas in (a) Isobaric process (p=constant) (b) Isothermal process (pV=constant) (c) Adiabatic process (`PV^gamma=` constant)
Answer» Correct Answer - A::D (a) Isobaric process `W=int_(V_i)^(V_f)pdV=p int_(V_i)^(V_f)dV` (as p= constant) `=p[V]_(V_i)^(V_f)=p(V_f-V_i)` `=pDeltaV` (b) Isothermal process `W=int_(V_i)^(V_f)pdV=int_(V_i)^(V_f)((nRT)/(V))dV` (as `p=(nRT)/(V)`) `=nRTint_(V_i)^(V_f)(dV)/(V)` (as T=constant) `=nRTIn((V_f)/(V_i))=nRTIn((p_i)/(p_f))` (as `p_iV_i=p_fV_f` So, `V_f/V_i=p_i/p_f`) (c) Adiabatic process `pV^gamma=` constant=k(say)=`p_iV_i^gamma=p_fV_f^gamma` Further, `p=(k)/(V^gamma)=kV^-gamma` `W=int_(V_i)^(V_f)pdV=int_(V_i)^(V_f)kV^(-gamma)dV=[(kV^(gamma+1))/(-gamma+1)]_(V_i)^(V_f)` `=(kV_f^(-gamma+1)-kV_i^(-gamma+1))/(-gamma+1)=(p_fV_f^gammaV_f^(-gamma+1)-p_iV_i^gammaV_i^(-gamma+1))/(1-gamma)` `=(p_fV_f-p_iV_i)/(1-gamma)=(nRT_f-nRT_i)/(1-gamma)=(nRDeltaT)/(1-gamma)`

Discussion

129.	In the process `pV^2=` constant, if temperature of gas is increased, thenA. (a) change in internal energy of gas is positiveB. (b) work done by gas is positiveC. (c) heat is given to the gasD. (d) heat is taken out from the gas
Answer» Correct Answer - A::C Temperature is increased. So, internal energy will also increase. `:.` `DeltaU=+ve` Further, `pV^2=` constant `:.` `((T)/(V))V^2=`constant or `Vprop1/T` Temperature is increased. So, volume will decrease and work done will be negative. In the process `pV^x`= constant, molar heat capacity is given by `C=C_V+(R)/(1-x)` Here, `x=2` `:.` `C=C_V-R` `C_V` of any gas is greater than R. So, C is positive. Hence, form the equation, `Q=nCDeltaT` Q is positive if T is increased. or `DeltaT` is positive.

Discussion

130.	The cyclic process form a circle on a pV diagram as shown in figure. The work done by the gas is A. (a) `pi/4(p_2-p_1)^2`B. (b) `pi/4(V_2-V_1)^2`C. (c) `pi/2(p_2-p_1)(V_2-V_1)`D. (d) `pi/4(p_2-p_1)(V_1-V_2)`
Answer» Correct Answer - D `W=-pi(Radius)^2` `=-pi(Radius)(Radius)` `=-pi((p_2-p_1)/(2))((V_2-V_1)/(2))`

Discussion

131.	Pressure and volume of a gas changes from `(p_0V_0)` to `(p_0/4, 2V_0)` in a process `pV^2=` constant. Find work done by the gas in the given process.
Answer» Correct Answer - A::B `pV^2=K=p_0V_0^2` `:.` `p=(K)/(V^2)` `W=intpdV=int_(V_0)^(2V_0)(K)/(V^2)dV` `=[-(K)/(V)]_(V_0)^(2V_0)` `=(K)/(V_0)-(K)/(2V_0)` `(K)/(2V_0)` Substituting, `K=p_0V_0^2`, we have `W=1/2p_0V_0`

Discussion

132.	A gas undergoes A to B through three different processes 1,2 and 3 as shown in the figure. The heat supplied to the gas is `Q_1`, `Q_2` and `Q_3` respectively, then A. (a) `Q_1=Q_2=Q_3`B. (b) `Q_1ltQ_2ltQ_3`C. (c) `Q_1gtQ_2gtQ_3`D. (d) `Q_1=Q_3gtQ_2`
Answer» Correct Answer - C `DeltaU` is same p. `Q=W+DeltaU` `W_1=+ve`, `W_2=0` and `W_3=-ve`

Discussion

133.	For an adiabatic compression the quantity pVA. (a) increasesB. (b) decreasesC. (c) remains constantD. (d) depends on `gamma`
Answer» Correct Answer - A In adiabatic compression temperature increases. Hence, pV increases, as `TproppV`

Discussion

134.	A system is given 300 calories of heat and it does 600 joules of work. How much does the internal energy of the system change in this process `(J=4.18 "joules"//cal)`A. 654 jouleB. 156.5 jouleC. `-300` jouleD. `-528.2` joule
Answer» Correct Answer - A `DeltaQ=DeltaU+DeltaW` `300xx4.18=DeltaU+600` `DeltaU=654J`

Discussion

135.	A monoatomic ideal gas, initially at temperature `T_1,` is enclosed in a cylinder fitted with a friction less piston. The gas is allowed to expand adiabatically to a temperature `T_2` by releasing the piston suddenly. If `L_1 and L_2` are the length of the gas column before expansion respectively, then `(T_1)/(T_2)` is given byA. (a) `((L_1)/(L_2))^(2/3)`B. (b) `L_1/L_2`C. (c) `L_2/L_1`D. (d) `((L_2)/(L_1))^(2/3)`
Answer» Correct Answer - D `TV^(gamma-1)=`constant `:.` `T_1/T_2=(V_2/V_1)^(gamma-1)` `=(L_2/L_1)^(5//3-1)=(L_2/L_1)^(2//3)`

Discussion

136.	The internal energy of a gas is given by `U=1.5pV. It expandeds from `100 cm^(3) to 200 cm^(3)` against a constant pressure of `1.0xx10^(5)` Pa. calculate the heat absorbed by the gas in the process.
Answer» Correct Answer - B Delta U = 1.5pV` `DeltaV = (20-100)cm^3 = 100cm^3` ` = 10^(-4)m^3` `P = 1xx10^5pa` `Delta U = 1.5xx10^5xx10^(-4) = 15J` `Delta W = 10^5xx10^(-4) = 10J` `DeltaQ = Delta U+DeltaW = 10+15 = 25J`

Discussion

137.	One of the most efficient engines ever developed operated between 2100K and 700K. Its actual efficiency is 40%. What percentage of its maximum possible efficiency is this?
Answer» `T_1=2100K`, `T_2=700K`, `eta_(actual)=40%` Maximum possible effiency of Carnot engine is `eta_(max)=1-(T_2)/(T_1)=1-700/2100` `=0.666=66.6%` Actual efficiency as the percentage of the maximum possible efficiency is `(eta_(actual))/(eta_(max))xx100=40/66.6xx100` `=60%`

Discussion

138.	At `27^@C` two moles of an ideal monatomic gas occupy a volume V. The gas expands adiabatically to a volume `2V`. Calculate (a) final temperature of the gas (b) change in its internal energy and (c) the work done by the gas during the process. [ `R=8.31J//mol-K`]
Answer» Correct Answer - A::B::C (a) In case of adiabatic change `TV^(gamma-1)=` constant So that `T_1V_1^(gamma-1)=T_2V_2^(gamma-1)` with `gamma=(5/3)` i.e. `300xxV^(2//3)=T(2V)^(2//3)` or `T=(300)/((2)^(2//3))=189K` (b) As `DeltaU=nC_VDeltaT=n(3/2R)DeltaT So, `DeltaU=2xx(3/2)xx8.31(189-300)` `=-2767.23J` Negative sign means internal energy will decrease. (c) According to first law of thermodynamics `Q=DeltaU+DeltaW` And as for adiabatic change `DeltaQ=0` `DeltaW=-DeltaU=2767.23J`

Discussion

139.	100 gram of ice at `0^@C` is converted into water vapour at `100^@C` Calculate the change in entropy.
Answer» `underset(0^(@)C)"ice"rarrunderset(0^(@)C)"Water"` `DeltaS_1=(DeltaQ)/(T_1)=(mL_(ice))/(T_1)=(100xx80)/(273)=29.3(cal)/(K)` `underset(0^(@)C)"Water"rarrunderset(100^(@)C)"Water"` `DeltaS_2=ms_(omega)ln(T_2)/(T_1)=100xx1xxln((373)/(273))` `=100ln(1.4)(cal)/(K)` `underset(100^(@)C)"Water"rarrunderset(100^(@)C)"Vapour"` `DeltaS=(DeltaQ)/(T_2)=(mL_(stream))/(T_2)=(100xx540)/(373)=144.8(cal)/(K)` Not change in entropy `DeltaS=DeltaS_1+DeltaS_2+DeltaS_3` `=29.3+100ln(1.4)+144.8` `=174.1+100ln(1.4)(cal)/(K)`

Discussion

140.	A sample of 100 g water is slowly heated from `27^@C` to `87^@C`. Calculate the change in the entropy of the water specific heat capacity of water `=4200(J)/(kg-K)`
Answer» `DeltaQ=msDeltaT` `DeltaS=(DeltaQ)/(T)=(msDeltaT)/(T)` The total change in entropy as temperature from `T_1` to `T_2`r `S_1-S_2=int_(T1)^(T2)ms(dT)/(T)` `=msln((T_2)/(T_1))` `=0.1xx4200ln((87+273)/(27+273))` `=420ln(1.2)(J)/(K)`

Discussion

141.	Find the change in the internal energy of 2 kg of water as it heated from `0^(0)C to 4^(0)C`. The specific heat capacity of water is `4200 J kg^(-1) K^(-1)` and its densities at `0^(0)C` and `4^(0)C` are `999.9 kg m^(-3)` and `1000 kg m^(-3)` respectively. atmospheric pressure `=10^(5)` Pa.
Answer» Correct Answer - B::C Given `M = 2kg` `Delta theta = 4^@C s_u = 4200J/kgK.` `p_0 = 999.9kg/m^3` `p_1 = 1000kg/m^3,P = 10^6kpa.` `Let internal energy = Delta V` `DeltaQ = DeltaU+DeltaW` `rArr ms Delta theta = DeltaU+P(V_0-V_4)` `rArr 2xx4200xx4 = delta U+10^2(V_0)-(V_4)` `rArr 33600 = DeltaU+10^5((m)/(p_0)-(m)/(p_4))` `rArr = Delta U +10^5xx0.0000002`ltbr.`rArr 33600 = Delta U +0.02` `Delta U = (33600-0.02)J`

Discussion

142.	Following figure shows two process A and B for a gas. If `Delta Q_(A)` and `Delta Q_(B)` are the amount of heat absorbed by the system in two case, and `Delta U_(A)` and `Delta U_(B)` are changes in internal energies, respectively, then : A. (a) `DeltaQ_1=Delta_2, DeltaU_1=DeltaU_2`B. (b) `DeltaQ_1gtDeltaQ_2,DeltaU_1gtDeltaU_2`C. (c) `DeltaQ_1ltDeltaQ_2,DeltaU_1ltDeltaU_2`D. (d) `DeltaQ_1gtDeltaQ_2,DeltaU_1=DeltaU_2`
Answer» Correct Answer - D `W_1gtW_2` as area under graph-`alpha` is more. Hence, `DeltaQ_1gtDeltaQ_2`, as `DeltaQ=W+DeltaU` and `DeltaU_1=DetlaU_2`

Discussion

143.	In a certain chemical process, a lab technician supplies 254 J of heat to a system. At the same time, 73 J of work are done on the system by its surroundings. What is the increase in the internal energy of the system?
Answer» Correct Answer - B::C `DeltaU=Q-W` `=254-(-73)=327J`

Discussion

144.	A diatomic ideal gas is heated at constant volume until its pressure becomes three times. It is again heated at constant pressure until its volume is doubled. Find the molar heat capacity for the whole process.
Answer» Correct Answer - A::C First process `V=const ant` `:.` `ppropT` Pressure becomes three times. Therefore, temperature also becomes three times. `(T_f=3T_i)` Second process `p=const ant` `:.` `VpropT` Volume is doubled. So, temperature also becomes two times. `(T_f=6T_i)` Now, `C=(Q)/(nDeltaT)=(Q_1+Q_2)/(nDeltaT)` `=(nC_VDeltaT_1+nC_pDeltaT_2)/(nDeltaT)` `=((5/2R)(3T_i-T_i)+(7/2R)(6T_i-3T_i))/(6T_i-T_i)` `=3.1R`

Discussion

145.	What work will be done, when 3 moles of an ideal gas are compreseed to half the initial volume at a constant temperature of 300K?A. `-5188 J`B. 5000JC. 5188JD. `-5000J`
Answer» Correct Answer - a As process is isothermal `W=2303nRT "log"((v_(2))/(v_(1)))` `W=2303xx3xx8.315xx300xx"log"((1)/(2))` `W=-5188J`

Discussion

146.	Find work done by the gas in the process AB shown in the following figures.
Answer» Correct Answer - B::C W = Area under p-V diagram Further `ppropV` along AB. Therefore, `p_B=2p_A=2p_0`

Discussion

147.	A heat enegine works on a carnot cycle with a heat sink at a temperature of `2700K`. The efficiency of the engine `10%`. Determine the temperature of heat source.
Answer» Efficiency of carnot Engine `eta=1-(T_2)/(T_1)` `(10)/(100)=1-(2700)/(T_1)` `(2700)/(T_1)=0.9` `T=3000K`

Discussion

148.	Which of the following is correct regarding adiabatic process (i) In adiabatic process, all the three variables, P, V and T changes (ii) In adiabatic process, the heat exchanged between system and surrounding is zero i.e., `DeltaQ=0` (iii) Since `DeltaQ=nCDeltaT`, therefore for adiabatic process molar specific heat `C=0` (iv) If a gas is suddenly expanded adiabatically temperature falls and if gas is adiabatically compressed, temperature risesA. (i),(iii)B. (ii),(iii)C. (iii),(iv)D. all
Answer» Correct Answer - D

Discussion

149.	`P-V` graph was obtained from state 1 to state 2 when a given mass of a gas is subjected to temperature changes during the process the gas isA. Heated continuouslyB. Cooled continuouslyC. Heated in the beginning and cooled towards the endD. cooled in the biginning and heated towards the end
Answer» Correct Answer - C

Discussion

150.	During an isothermal expansion of an ideal gasA. Its internal energy decreasesB. Its internal energy does not changeC. the work done by the gas is equal to the quantity of heat absorbed by it.D. both b and c are correct
Answer» Correct Answer - d In isothermal process, temperautre remains constant `because DeltaU=nC_(v)DeltaT=0 (because DeltaT=0)` According to the first law of thermodynamics. `DeltaH=DeltaU+DeltaW` `therefore DeltaH=DeltaW (because DeltaU=0)` Hence, (d) is correct.

Discussion

Explore topic-wise InterviewSolutions in .

An ideal gas is initially at temperature T and volume V. Its volume is increased by `DeltaV` due to an increase in temperature `DeltaT,` pressure remaining constant. The quantity `delta=(DeltaV)/(VDeltaT)` varies with temperature asA. B. C. D.

As shown in figure three paths through which a gas can be taken from the state A to the state B. Calculate the work done by the gas in each of the three paths.

For an ideal gas the molar heat capacity varies as `C=C_V+3aT^2`. Find the equation of the process in the variables (T,V) where a is a constant.

The molar heat capacity for an ideal gas cannotA. cannot be negativeB. must be equal to either `C_v` or `C_p`C. must lie in the range `C_vleCleC-p`D. may have any value between `-infty` and `+infty`

In the cyclic process shown in the `V-P` diagram the magnitude of the work is done isA. `pi((P_1-P_2)/(2))^2`B. `pi((V_1-V_2)/(2))^2`C. `(pi)/(4)(P_2-P_1)(V_2-V_1)`D. `pi(P_2V_2-P_1V_1)`

Let `(C_v) and (C_p)` denote the molar heat capacities of an ideal gas at constant volume and constant pressure respectively . Which of the following is a universal constant?A. `(C_p)/(C_v)`B. `C_pC-v`C. `C_p-C_v`D. `C_p+C_v`

Each molecule of a gas has `f` degrees of freedom. The ratio `gamma` for the gas isA. `1+(f)/(2)`B. `1+(1)/(f)`C. `1+(2)/(f)`D. `1+((f-1))/(3)`

A cornot engine in 1000 kilocal of heat from a reservoir at `627^@C` and exhausts it to sink at `27^@C`. What is its efficiency? How much work does it perform?

A Carnot engine takes `3xx10^6cal`. of heat from a reservoir at `62^@C`, and gives it to a sink at `27^@C`. The work done by the engine is

A rigid container of negligible heat capacity contains one mole of an ideal gas. The temperatures of the gas increases by `1^@C` if `3.0` cal of heat is added to it. The gas may beA. (i),(ii)B. (ii),(iii)C. (iii),(iv)D. (i),(iv)

An ideal gas goes from the state `i` to the state `f` as shown in the figure.The work done by the gas during the process A. is positiveB. is negativeC. is zeroD. none

In an adiabatic change, the pressure p and temperature T of a diatomic gas are related by the relation `ppropT^alpha`, where `alpha` equalsA. (a) 1.67B. (b) 0.4C. (c) 0.6D. (d) 3.5

The adiabatic elasticity of hydrogen gas `(gamma=1.4)` at `NTP`A. `1xx10^5(N)/(m^2)`B. `1xx10^8(N)/(m^2)`C. `1.4(N)/(m^2)`D. `1.4xx10^5(N)/(m^2)`

In an adiabatic change, the pressure p and temperature T of a diatomic gas are related by the relation `ppropT^alpha`, where `alpha` equalsA. `(5)/(3)`B. `(2)/(5)`C. `(3)/(5)`D. `(5)/(2)`

The work of 146 kJ is performed in order to compress one kilo mole of a gas adiabatically and in this process the temperature of the gas increases by `7^@C`. The gas is `(R=8.3ml^-1Jmol^-1K^-1)`A. TriatomicB. A mixture of monoatomic and idatomicC. MonoatomicD. Diatomic

A gas undergoes a process in which its pressure `P` and volume `V` are related as `VP^(n) =` constant. The bulk modulus of the gas in the process is:A. `np`B. `p^((1)/(n))`C. `(p)/(n)`D. `p^n`

During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio `C_P//C_V` for the gas isA. 2B. 1.5C. `(5)/(3)`D. `(4)/(3)`

A gas follows a process `TV^(n-1)=const ant`, where `T=` absolute temperature of the gas and `V=` volume of the gas. The bulk modulus of the gas in the process is given byA. (a) `(n-1)p`B. (b) `p//(n-1)`C. (c) `np`D. (d) `p//n`

An ideal gas expands while the pressure is kept constant. During the process, does heat flow into the gas or out of the gas? Justify you answer.

A sample of ideal gas is expanded to twice its original volume of `1.00m^3` in a quasi-static process for which `p=alphaV^2`, with `alpha=5.00 atm//m^6`, as shown in Fig. How much work is done by the expanding gas?

Show how internal energy U varies with T in isochoric, isobaric and adiabatic process?

Refer to figure in previous question, `DeltaU_(1)` and `DeltaU_(2)` be the changes in internal energy of the system in the processes `A` and `B` thenA. `DeltaU_1 gt DeltaU_2`B. `DeltaU_1=DeltaU_2`C. `DeltaU_1 lt DeltaU_2`D. `DeltaU_1neDeltaU_2`

Refer to let `DeltaU_(1)` and `DeltaU_(2)` be the changes in internal energy of the system in the processes A and B. thenA. `DeltaU_(1)gtDeltaU_(2)`B. `DeltaU_(1)=DeltaU_(2)`C. `DeltaU_(1)ltDeltaU_(2)`D. `DeltaU_(1)!=DeltaU_(2)`

Method 1 of W By integration, make expressions of work done by gas in (a) Isobaric process (p=constant) (b) Isothermal process (pV=constant) (c) Adiabatic process (`PV^gamma=` constant)

In the process `pV^2=` constant, if temperature of gas is increased, thenA. (a) change in internal energy of gas is positiveB. (b) work done by gas is positiveC. (c) heat is given to the gasD. (d) heat is taken out from the gas

The cyclic process form a circle on a pV diagram as shown in figure. The work done by the gas is A. (a) `pi/4(p_2-p_1)^2`B. (b) `pi/4(V_2-V_1)^2`C. (c) `pi/2(p_2-p_1)(V_2-V_1)`D. (d) `pi/4(p_2-p_1)(V_1-V_2)`

Pressure and volume of a gas changes from `(p_0V_0)` to `(p_0/4, 2V_0)` in a process `pV^2=` constant. Find work done by the gas in the given process.

A gas undergoes A to B through three different processes 1,2 and 3 as shown in the figure. The heat supplied to the gas is `Q_1`, `Q_2` and `Q_3` respectively, then A. (a) `Q_1=Q_2=Q_3`B. (b) `Q_1ltQ_2ltQ_3`C. (c) `Q_1gtQ_2gtQ_3`D. (d) `Q_1=Q_3gtQ_2`

For an adiabatic compression the quantity pVA. (a) increasesB. (b) decreasesC. (c) remains constantD. (d) depends on `gamma`

A system is given 300 calories of heat and it does 600 joules of work. How much does the internal energy of the system change in this process `(J=4.18 "joules"//cal)`A. 654 jouleB. 156.5 jouleC. `-300` jouleD. `-528.2` joule

The internal energy of a gas is given by `U=1.5pV. It expandeds from `100 cm^(3) to 200 cm^(3)` against a constant pressure of `1.0xx10^(5)` Pa. calculate the heat absorbed by the gas in the process.

One of the most efficient engines ever developed operated between 2100K and 700K. Its actual efficiency is 40%. What percentage of its maximum possible efficiency is this?

At `27^@C` two moles of an ideal monatomic gas occupy a volume V. The gas expands adiabatically to a volume `2V`. Calculate (a) final temperature of the gas (b) change in its internal energy and (c) the work done by the gas during the process. [ `R=8.31J//mol-K`]

100 gram of ice at `0^@C` is converted into water vapour at `100^@C` Calculate the change in entropy.

A sample of 100 g water is slowly heated from `27^@C` to `87^@C`. Calculate the change in the entropy of the water specific heat capacity of water `=4200(J)/(kg-K)`

Find the change in the internal energy of 2 kg of water as it heated from `0^(0)C to 4^(0)C`. The specific heat capacity of water is `4200 J kg^(-1) K^(-1)` and its densities at `0^(0)C` and `4^(0)C` are `999.9 kg m^(-3)` and `1000 kg m^(-3)` respectively. atmospheric pressure `=10^(5)` Pa.

In a certain chemical process, a lab technician supplies 254 J of heat to a system. At the same time, 73 J of work are done on the system by its surroundings. What is the increase in the internal energy of the system?

A diatomic ideal gas is heated at constant volume until its pressure becomes three times. It is again heated at constant pressure until its volume is doubled. Find the molar heat capacity for the whole process.

What work will be done, when 3 moles of an ideal gas are compreseed to half the initial volume at a constant temperature of 300K?A. `-5188 J`B. 5000JC. 5188JD. `-5000J`

Find work done by the gas in the process AB shown in the following figures.

A heat enegine works on a carnot cycle with a heat sink at a temperature of `2700K`. The efficiency of the engine `10%`. Determine the temperature of heat source.

`P-V` graph was obtained from state 1 to state 2 when a given mass of a gas is subjected to temperature changes during the process the gas isA. Heated continuouslyB. Cooled continuouslyC. Heated in the beginning and cooled towards the endD. cooled in the biginning and heated towards the end

During an isothermal expansion of an ideal gasA. Its internal energy decreasesB. Its internal energy does not changeC. the work done by the gas is equal to the quantity of heat absorbed by it.D. both b and c are correct