The expression for entropy change given by, ΔS = nR ln (V2/V1) + nCvln (T2/T1) is valid for (A) Reversible isothermal volume change (B) Heating of a substance (C) Cooling of a substance (D) Simultaneous heating and expansion of an ideal gas

The expression for entropy change given by, ΔS = nR ln (V2/V1) + nCvln (T2/T1) is valid for
(A) Reversible isothermal volume change
(B) Heating of a substance
(C) Cooling of a substance
(D) Simultaneous heating and expansion of an ideal gas
by

1 Answer

Answer :

(D) Simultaneous heating and expansion of an ideal gas
by

Related questions

The expression for entropy change given by, ΔS = - nR ln (P2/P1), holds good for (A) Expansion of a real gas (B) Reversible isothermal volume change (C) Heating of an ideal gas (D) Cooling of a real gas

Description : The expression for entropy change given by, ΔS = - nR ln (P2/P1), holds good for (A) Expansion of a real gas (B) Reversible isothermal volume change (C) Heating of an ideal gas (D) Cooling of a real gas

Last Answer : (B) Reversible isothermal volume change

The expression for entropy change, ΔS = n Cp. ln (T2/T1), is valid for the __________ of a substance. (A) Simultaneous pressure & temperature change (B) Heating (C) Cooling (D) Both (B) and (C)

Description : The expression for entropy change, ΔS = n Cp. ln (T2/T1), is valid for the __________ of a substance. (A) Simultaneous pressure & temperature change (B) Heating (C) Cooling (D) Both (B) and (C)

Last Answer : (D) Both (B) and (C)

The expression, ∆G = nRT. ln(P2/P1), gives the free energy change (A) With pressure changes at constant temperature (B) Under reversible isothermal volume change (C) During heating of an ideal gas (D) During cooling of an ideal gas

Description : The expression, ∆G = nRT. ln(P2/P1), gives the free energy change (A) With pressure changes at constant temperature (B) Under reversible isothermal volume change (C) During heating of an ideal gas (D) During cooling of an ideal gas

Last Answer : (A) With pressure changes at constant temperature

What is the equation for the work done by a constant temperature system?  A. W = mRTln(V2-V1)  B. W = mR( T2-T1 ) ln( V2/V1)  C. W = mRTln (V2/V1)  D. W = RT ln (V2/V1) Formula : W=∫ pdV lim1,2 ∫ = mRT / V

Description : What is the equation for the work done by a constant temperature system?  A. W = mRTln(V2-V1)  B. W = mR( T2-T1 ) ln( V2/V1)  C. W = mRTln (V2/V1)  D. W = RT ln (V2/V1) Formula : W=∫ pdV lim1,2 ∫ = mRT / V

Last Answer : W = mRTln (V2/V1)

A certain gas with cp = 0.529Btu/lb°R and R = 96.2ft/lbºR expands from 5 ft and 80ºF to 15 ft while the pressure remains constant at 15.5 psia.  a. T2=1.620ºR, ∫H = 122.83 Btu  b. T2 = 2°R, ∫H = 122.83 Btu  c. T2 = 2.620ºR, ∫H = 122.83 Btu  d. T2 = 1°R, ∫H = 122.83 Btu T2= V2(t2)/V1 and ∫H = mcp (T2-T1)

Description : A certain gas with cp = 0.529Btu/lb°R and R = 96.2ft/lbºR expands from 5 ft and 80ºF to 15 ft while the pressure remains constant at 15.5 psia.  a. T2=1.620ºR, ∫H = 122.83 Btu  b. T2 = 2°R, ∫H = 122.83 Btu  c. ... , ∫H = 122.83 Btu  d. T2 = 1°R, ∫H = 122.83 Btu T2= V2(t2)/V1 and ∫H = mcp (T2-T1)

Last Answer : T2=1.620ºR, ∫H = 122.83 Btu

To obtain integrated form of Clausius-Clapeyron equation, ln (P2/P1) = (∆HV/R) (1/T1- 1/T2) from the exact Clapeyron equation, it is assumed that the (A) Volume of the liquid phase is negligible compared to that of vapour phase (B) Vapour phase behaves as an ideal gas (C) Heat of vaporisation is independent of temperature (D) All (A), (B) & (C)

Description : To obtain integrated form of Clausius-Clapeyron equation, ln (P2/P1) = (∆HV/R) (1/T1- 1/T2) from the exact Clapeyron equation, it is assumed that the (A) Volume of the liquid phase is negligible compared to ... gas (C) Heat of vaporisation is independent of temperature (D) All (A), (B) & (C)

Last Answer : (D) All (A), (B) & (C)

Which of the following is the mathematical representation of the Charles’s law?  A. V1/V2= P2/P1  B. V1/T1=V2/T2  C. V1/T2=V2/T1  D. V1/V2=√P2/√P1

Description : Which of the following is the mathematical representation of the Charles’s law?  A. V1/V2= P2/P1  B. V1/T1=V2/T2  C. V1/T2=V2/T1  D. V1/V2=√P2/√P1

Last Answer : V1/T1=V2/T2

The expression, nCv(T2- T1), is for the __________ of an ideal gas. (A) Work done under adiabatic condition (B) Co-efficient of thermal expansion (C) Compressibility (D) None of these

Description : The expression, nCv(T2- T1), is for the __________ of an ideal gas. (A) Work done under adiabatic condition (B) Co-efficient of thermal expansion (C) Compressibility (D) None of these

Last Answer : (A) Work done under adiabatic condition

The expression, nRT ln(P1/P2), is for the __________of an ideal gas. (A) Compressibility (B) Work done under adiabatic condition (C) Work done under isothermal condition (D) Co-efficient of thermal expansion

Description : The expression, nRT ln(P1/P2), is for the __________of an ideal gas. (A) Compressibility (B) Work done under adiabatic condition (C) Work done under isothermal condition (D) Co-efficient of thermal expansion

Last Answer : C) Work done under isothermal condition

In a totally irreversible isothermal expansion process for an ideal gas, ΔE = 0, ΔH = 0. Then ΔQ and ΔS will be (A) ΔQ = 0, ΔS=0 (B) ΔQ = 0, ΔS = +ve (C) ΔQ = 0, ΔS = -ve (D) ΔQ = +ve, ΔS= +ve

Description : In a totally irreversible isothermal expansion process for an ideal gas, ΔE = 0, ΔH = 0. Then ΔQ and ΔS will be (A) ΔQ = 0, ΔS=0 (B) ΔQ = 0, ΔS = +ve (C) ΔQ = 0, ΔS = -ve (D) ΔQ = +ve, ΔS= +ve

Last Answer : (B) ΔQ = 0, ΔS = +ve

. The entropy change in a reversible isothermal process, when an ideal gas expands to four times its initial volume is (A) R loge 4 (B) R log10 4 (C) Cv log10 4 (D) Cv loge 4

Description : . The entropy change in a reversible isothermal process, when an ideal gas expands to four times its initial volume is (A) R loge 4 (B) R log10 4 (C) Cv log10 4 (D) Cv loge 4

Last Answer : (A) R loge 4

. The entropy change in a reversible isothermal process, when an ideal gas expands to four times its initial volume is (A) R loge 4 (B) R log10 4 (C) Cv log10 4 (D) Cv loge 4

Description : . The entropy change in a reversible isothermal process, when an ideal gas expands to four times its initial volume is (A) R loge 4 (B) R log10 4 (C) Cv log10 4 (D) Cv loge 4

Last Answer : (A) R loge 4

The compressibility factor of a gas is given by (where, V1 = actual volume of the gas V2 = gas volume predicted by ideal gas law) (A) V1/V2 (B) V2/V1 (C) V1- V2 (D) V1.V2

Description : The compressibility factor of a gas is given by (where, V1 = actual volume of the gas V2 = gas volume predicted by ideal gas law) (A) V1/V2 (B) V2/V1 (C) V1- V2 (D) V1.V2

Last Answer : (A) V1/V2

Helium ( R= 0.4698 BTU/lbm-˚R ) is compressed isothermally from 14.7 psia and 68 ˚F. The compression ratio is 1:4. Calculate the work done by the gas.  A. –1454 BTU/lbm  B. -364 BTU/lbm  C.-187BTU/lbm  D.46.7 BTU/lbm Formula: W = RT ln (V2/V1)

Description : Helium ( R= 0.4698 BTU/lbm-˚R ) is compressed isothermally from 14.7 psia and 68 ˚F. The compression ratio is 1:4. Calculate the work done by the gas.  A. –1454 BTU/lbm  B. -364 BTU/lbm  C.-187BTU/lbm  D.46.7 BTU/lbm Formula: W = RT ln (V2/V1)

Last Answer : -364 BTU/lbm

For the reversible isothermal expansion of one mole of an ideal gas at 300 K, from a volume of `10 dm^(3)` to `20 dm^(3), Delta H` is -

Description : For the reversible isothermal expansion of one mole of an ideal gas at 300 K, from a volume of `10 dm^(3)` to `20 dm^(3), Delta H` is -

Last Answer : For the reversible isothermal expansion of one mole of an ideal gas at 300 K, from a volume of `10 dm^(3)` to ` ... . `-1.73 KJ` C. `3.46 KJ` D. Zero

Entropy of a substance remains constant during a/an __________ change. (A) Reversible isothermal (B) Irreversible isothermal (C) Reversible adiabatic (D) None of these

Description : Entropy of a substance remains constant during a/an __________ change. (A) Reversible isothermal (B) Irreversible isothermal (C) Reversible adiabatic (D) None of these

Last Answer : (C) Reversible adiabatic

Gas is enclosed in a cylinder with a weighted piston as the stop boundary. The gas is heated and expands from a volume of 0.04 m^3 to 0.10 m^3 at a constant pressure of 200kPa.Calculate the work done by the system.  A. 8 kJ  B. 10 kJ  C.12 kJ  D.14 kJ Formula: W = p(V2-V1)

Description : Gas is enclosed in a cylinder with a weighted piston as the stop boundary. The gas is heated and expands from a volume of 0.04 m^3 to 0.10 m^3 at a constant pressure of 200kPa.Calculate the work done by the system.  A. 8 kJ  B. 10 kJ  C.12 kJ  D.14 kJ Formula: W = p(V2-V1)

Last Answer : 12 kJ

Twenty grams of oxygen gas are compressed at a constant temperature of 30 ˚C to 5%of their original volume. What work is done on the system.  A.824 cal  B.924 cal  C.944 cal  D.1124 cal Formula: W = -mRTln (V2/V1) Where R = (1.98 cal/gmole·K) (32 g/gmole)

Description : Twenty grams of oxygen gas are compressed at a constant temperature of 30 ˚C to 5%of their original volume. What work is done on the system.  A.824 cal  B.924 cal  C.944 cal  D.1124 cal Formula: W = -mRTln (V2/V1) Where R = (1.98 cal/gmole·K) (32 g/gmole)

Last Answer : 1124 cal

An ideal gas at 45psig and 80ºF is heated in the close container to 130ºF. What is the final pressure?  a. 65.10 psi  b. 65.11 psi  c. 65.23 psi  d. 61.16 psi P1V1/T1= P2V2/T2;V = Constant

Description : An ideal gas at 45psig and 80ºF is heated in the close container to 130ºF. What is the final pressure?  a. 65.10 psi  b. 65.11 psi  c. 65.23 psi  d. 61.16 psi P1V1/T1= P2V2/T2;V = Constant

Last Answer : 65.23 psi

Which of the following is the Ideal gas law (equation)?  A. V/T = K  B. V= k*(1/P)  C. P1/T1 = P2/T2  D. PV = nRT

Description : Which of the following is the Ideal gas law (equation)?  A. V/T = K  B. V= k*(1/P)  C. P1/T1 = P2/T2  D. PV = nRT

Last Answer : PV = nRT

Pick out the wrong statement: (A) The expansion of a gas in vacuum is an irreversible process (B) An isometric process is a constant pressure process (C) Entropy change for a reversible adiabatic process is zero (D) Free energy change for a spontaneous process is negative

Description : Pick out the wrong statement: (A) The expansion of a gas in vacuum is an irreversible process (B) An isometric process is a constant pressure process (C) Entropy change for a reversible adiabatic process is zero (D) Free energy change for a spontaneous process is negative

Last Answer : (B) An isometric process is a constant pressure process

What happens in a reversible adiabatic expansion process? (A) Heating takes place (B) Cooling takes place (C) Pressure is constant (D) Temperature is constant

Description : What happens in a reversible adiabatic expansion process? (A) Heating takes place (B) Cooling takes place (C) Pressure is constant (D) Temperature is constant

Last Answer : (B) Cooling takes place

Which of the following processes are thermodynamically reversible  (a) throttling  (b) free expansion  (c) constant volume and constant pressure  (d) hyperbolic and pV = C  (e) isothermal and adiabatic.

Description : Which of the following processes are thermodynamically reversible  (a) throttling  (b) free expansion  (c) constant volume and constant pressure  (d) hyperbolic and pV = C  (e) isothermal and adiabatic.

Last Answer : Answer : e

Q No: 251 In a sludge digestion tank if the moisture content of sludge V1 litres is reduced from p1 % to p2 % the volume V2 is A. [(100 + P1)/(100 – P2)] V1 B. [(100 – P1)/(100 + P2)] V1 C. [(100 – P1)/(100 – P2)] V1 D. [(100 + P2)/(100 – P1)] V1

Description : Q No: 251 In a sludge digestion tank if the moisture content of sludge V1 litres is reduced from p1 % to p2 % the volume V2 is A. [(100 + P1)/(100 – P2)] V1 B. [(100 – P1)/(100 + P2)] V1 C. [(100 – P1)/(100 – P2)] V1 D. [(100 + P2)/(100 – P1)] V1

Last Answer : ANS: C

Assuming compression is according to the Law PV = C, Calculate the initial volume of the gas at a pressure of 2 bars w/c will occupy a volume of 6m³ when it is compressed to a pressure of 42 Bars.  a) 130m³  b) 136m³  c) 120m³  d) 126m³ Formula: P1V1/T1 =P2V2/T2

Description : Assuming compression is according to the Law PV = C, Calculate the initial volume of the gas at a pressure of 2 bars w/c will occupy a volume of 6m³ when it is compressed to a pressure of 42 Bars.  a) 130m³  b) 136m³  c) 120m³  d) 126m³ Formula: P1V1/T1 =P2V2/T2

Last Answer : 126m³

A gas having a volume of100 ft³ at 27ºC is expanded to 120 ft³by heated at constant pressure to what temperature has it been heated to have this new volume?  a. 87°C  b. 85°C  c. 76°C  d. 97°C t2= T2–T1

Description : A gas having a volume of100 ft³ at 27ºC is expanded to 120 ft³by heated at constant pressure to what temperature has it been heated to have this new volume?  a. 87°C  b. 85°C  c. 76°C  d. 97°C t2= T2–T1

Last Answer : 87°C

The volume of the gas held at constant pressure increases 4 cm² at 0°C to 5cm². What is the final pressure?  a. 68.65ºC  b. 68.25ºC  c. 70.01°C  d. 79.1ºC t2= T2–T1

Description : The volume of the gas held at constant pressure increases 4 cm² at 0°C to 5cm². What is the final pressure?  a. 68.65ºC  b. 68.25ºC  c. 70.01°C  d. 79.1ºC t2= T2–T1

Last Answer : 981 N

There are 1.36 kg of gas, for which R = 377 J/kg.k and k = 1.25, that undergo a nonflow constant volume process from p1 = 551.6 kPa and t1 = 60°C to p2 = 1655 kPa. During the process the gas is internally stirred and there are also added 105.5 kJ of heat. Determine t2. (Formula: T2= T1p2/ p1)  a. 999 K  b. 888 K  c. 456 K  d. One of the above

Description : There are 1.36 kg of gas, for which R = 377 J/kg.k and k = 1.25, that undergo a nonflow constant volume process from p1 = 551.6 kPa and t1 = 60°C to p2 = 1655 kPa. During the process the gas is internally stirred and ... (Formula: T2= T1p2/ p1)  a. 999 K  b. 888 K  c. 456 K  d. One of the above

Last Answer : 999 K

The heat supplied to the gaS at constant volume is (where m = Mass of gas, cv = Specific heat at constant volume, cp = Specific heat at constant pressure, T2 – T1 = Rise in temperature, and R = Gas constant)  A. mR(T2 – T1)  B. mcv(T2 – T1)  C. mcp(T2 – T1)  D. mcp(T2 + T1)

Description : The heat supplied to the gaS at constant volume is (where m = Mass of gas, cv = Specific heat at constant volume, cp = Specific heat at constant pressure, T2 – T1 = Rise in temperature, and R = Gas constant)  A. mR(T2 – T1)  B. mcv(T2 – T1)  C. mcp(T2 – T1)  D. mcp(T2 + T1)

Last Answer : Answer: B

High pressure steam is expanded adiabatically and reversibly through a well insulated turbine, which produces some shaft work. If the enthalpy change and entropy change across the turbine are represented by ΔH and ΔS respectively for this process: (A) Δ H = 0 and ΔS = 0 (B) Δ H ≠ 0 and ΔS = 0 (C) Δ H ≠ 0 and ΔS ≠ 0 (D) Δ H = 0 and ΔS ≠ 0

Description : High pressure steam is expanded adiabatically and reversibly through a well insulated turbine, which produces some shaft work. If the enthalpy change and entropy change across the turbine are represented by ΔH and ΔS respectively for this process: ... (C) Δ H ≠ 0 and ΔS ≠ 0 (D) Δ H = 0 and ΔS ≠ 0

Last Answer : (B) Δ H ≠ 0 and ΔS = 0

The workdone in ergs for the reversible expansion of one mole of an ideal gas from a volume of 10 litres to 20 litres at `25^(@)C` is

Description : The workdone in ergs for the reversible expansion of one mole of an ideal gas from a volume of 10 litres to 20 litres at `25^(@)C` is

Last Answer : The workdone in ergs for the reversible expansion of one mole of an ideal gas from a volume of 10 ... 0821xx298xx` log 0.5 D. `-2.303xx2xx298` log2

Throttling process is a/an __________ process. (A) Reversible and isothermal (B) Irreversible and constant enthalpy (C) Reversible and constant entropy (D) Reversible and constant enthalpy

Description : Throttling process is a/an __________ process. (A) Reversible and isothermal (B) Irreversible and constant enthalpy (C) Reversible and constant entropy (D) Reversible and constant enthalpy

Last Answer : (B) Irreversible and constant enthalpy

On a P-V diagram of an ideal gas, suppose a reversible adiabatic line intersects a reversible isothermal line at point A. Then at a point A, the slope of the reversible adiabatic line (∂P/∂V)s and the slope of the reversible isothermal line (∂P/∂V)T are related as (where, y = Cp/Cv) (A) (∂P/∂V)S = (∂P/∂V)T (B) (∂P/∂V)S = [(∂P/∂V)T]Y (C) (∂P/∂V)S = y(∂P/∂V)T (D) (∂P/∂V)S = 1/y(∂P/∂V)T

Description : On a P-V diagram of an ideal gas, suppose a reversible adiabatic line intersects a reversible isothermal line at point A. Then at a point A, the slope of the reversible adiabatic line (∂P/∂V)s and the slope of the reversible isothermal line ... Y (C) (∂P/∂V)S = y(∂P/∂V)T (D) (∂P/∂V)S = 1/y(∂P/∂V)T

Last Answer : (C) (∂P/∂V)S = y(∂P/∂V)T

For an isothermal reversible compression of an ideal gas (A) Only ΔE = 0 (B) Only ΔH =0 (C) ΔE = ΔH = 0 (D) dQ = dE

Description : For an isothermal reversible compression of an ideal gas (A) Only ΔE = 0 (B) Only ΔH =0 (C) ΔE = ΔH = 0 (D) dQ = dE

Last Answer : C) ΔE = ΔH = 0

When the expansion of compression of gas takes place without transfer of heat to or from the gas the process is called  a. reversible  b. adiabatic  c. polytropic  d. isothermal

Description : When the expansion of compression of gas takes place without transfer of heat to or from the gas the process is called  a. reversible  b. adiabatic  c. polytropic  d. isothermal

Last Answer : adiabatic

Pick out the correct statement. (A) Like internal energy and enthalpy, the absolute value of standard entropy for elementary substances is zero (B) Melting of ice involves increase in enthalpy and a decrease in randomness (C) The internal energy of an ideal gas depends only on its pressure (D) Maximum work is done under reversible conditions

Description : Pick out the correct statement. (A) Like internal energy and enthalpy, the absolute value of standard entropy for elementary substances is zero (B) Melting of ice involves increase in enthalpy and ... of an ideal gas depends only on its pressure (D) Maximum work is done under reversible conditions

Last Answer : (D) Maximum work is done under reversible conditions

In case of a reversible process (following pvn = constant), work obtained for trebling the volume (v1 = 1 m3 and v23 m3) is maximum, when the value of 'n' is (A) 0 (B) 1 (C) y = 1.44 (D) 1.66

Description : In case of a reversible process (following pvn = constant), work obtained for trebling the volume (v1 = 1 m3 and v23 m3) is maximum, when the value of 'n' is (A) 0 (B) 1 (C) y = 1.44 (D) 1.66

Last Answer : (A) 0

Ten cu. ft of air at 300psia and 400°F is cooled to 140°F at constant volume. What is the transferred heat?  a.-120Btu  b. -220Btu  c.-320Btu  d. -420Btu formula: Q= mcv(T2-T1)

Description : Ten cu. ft of air at 300psia and 400°F is cooled to 140°F at constant volume. What is the transferred heat?  a.-120Btu  b. -220Btu  c.-320Btu  d. -420Btu formula: Q= mcv(T2-T1)

Last Answer : -420Btu

What is the resulting pressure when one pound of air at 15 psia and 200 ˚F is heated at constant volume to 800 ˚F?  A.15 psia  B. 28.6 psia  C. 36.4 psia.  D. 52.1 psia Formula : T1/p1 = T2/p2 p2= p1T2 / T1

Description : What is the resulting pressure when one pound of air at 15 psia and 200 ˚F is heated at constant volume to 800 ˚F?  A.15 psia  B. 28.6 psia  C. 36.4 psia.  D. 52.1 psia Formula : T1/p1 = T2/p2 p2= p1T2 / T1

Last Answer : 28.6 psia

Consider an ideal solution of components A and B. The entropy of mixing per mole of an alloy containing 50% B is (A) R ln 2 (B) -R ln 2 (C) 3 R ln 2 (D) -3R ln 2

Description : Consider an ideal solution of components A and B. The entropy of mixing per mole of an alloy containing 50% B is (A) R ln 2 (B) -R ln 2 (C) 3 R ln 2 (D) -3R ln 2

Last Answer : (A) R ln 2

essure drop (Δp) for a fluid flowing in turbulent flow through a pipe is a function of velocity (V) as (A) V1.8 (B) V-0.2 (C) V2.7 (D) V

Description : essure drop (Δp) for a fluid flowing in turbulent flow through a pipe is a function of velocity (V) as (A) V1.8 (B) V-0.2 (C) V2.7 (D) V

Last Answer : (D) V

A certain gas, with cp = 0.529Btu/ lb. °Rand R = 96.2ft.lb/lb. °R, expands from 5 cu ft and 80°F to 15 cu ft while the pressure remains constant at 15.5psia. Compute for T2.  a.1520°R  b. 1620°R  c. 1720°R  d. 1820°R formula: T2= T1V2/V1

Description : A certain gas, with cp = 0.529Btu/ lb. °Rand R = 96.2ft.lb/lb. °R, expands from 5 cu ft and 80°F to 15 cu ft while the pressure remains constant at 15.5psia. Compute for T2.  a.1520°R  b. 1620°R  c. 1720°R  d. 1820°R formula: T2= T1V2/V1

Last Answer : 1620°R

A certain gas, with cp = 0.529Btu/lb.°R and R = 96.2 ft.lb/lb.°R, expands from 5 cu ft and 80°F to 15 cu ft while the pressure remains constant at 15.5 psia. Compute for T2. (Formula: T2= T1V2/V1)  a. 460°R  b. 270°R  c. 1620 °R  d. None of the above

Description : A certain gas, with cp = 0.529Btu/lb.°R and R = 96.2 ft.lb/lb.°R, expands from 5 cu ft and 80°F to 15 cu ft while the pressure remains constant at 15.5 psia. Compute for T2. (Formula: T2= T1V2/V1)  a. 460°R  b. 270°R  c. 1620 °R  d. None of the above

Last Answer : 1620 °R

Pick out the wrong statement. (A) The net change in entropy in any reversible cycle is always zero (B) The entropy of the system as a whole in an irreversible process increases (C) The entropy of the universe tends to a maximum (D) The entropy of a substance does not remain constant during a reversible adiabatic change

Description : Pick out the wrong statement. (A) The net change in entropy in any reversible cycle is always zero (B) The entropy of the system as a whole in an irreversible process increases (C) The ... to a maximum (D) The entropy of a substance does not remain constant during a reversible adiabatic change

Last Answer : (D) The entropy of a substance does not remain constant during a reversible adiabatic change

The efficiency of a Carnot heat engine operating between absolute temperatures T1 and T2 (when, T1 > T2) is given by (T1- T2)/T1. The co efficient of performance (C.O.P.) of a Carnot heat pump operating between T1 and T2is given by (A) T1/(T1-T2) (B) T2/(T1-T2) (C) T1/T2 (D) T2/R1

Description : The efficiency of a Carnot heat engine operating between absolute temperatures T1 and T2 (when, T1 > T2) is given by (T1- T2)/T1. The co efficient of performance (C.O.P.) of a Carnot heat pump operating between T1 and T2is given by (A) T1/(T1-T2) (B) T2/(T1-T2) (C) T1/T2 (D) T2/R1

Last Answer : (A) T1/(T1-T2)

The ratio of equilibrium constants (Kp2/Kp1) at two different temperatures is given by (A) (R/∆H) (1/T1- 1/T2) (B) (∆H/R) (1/T1- 1/T2) (C) (∆H/R) (1/T2- 1/T1) (D) (1/R) (1/T1- 1/T2)

Description : The ratio of equilibrium constants (Kp2/Kp1) at two different temperatures is given by (A) (R/∆H) (1/T1- 1/T2) (B) (∆H/R) (1/T1- 1/T2) (C) (∆H/R) (1/T2- 1/T1) (D) (1/R) (1/T1- 1/T2)

Last Answer : (B) (∆H/R) (1/T1- 1/T2)

The equilibrium constant for a chemical reaction at two different temperatures is given by (A) Kp2/Kp1 = - (∆H/R) (1/T2- 1/T1) (B) Kp2/Kp1 = (∆H/R) (1/T2- 1/T1) (C) Kp2/Kp1 = ∆H (1/T2- 1/T1) (D) Kp2/Kp1 = - (1/R) (1/T2- 1/T1)

Description : The equilibrium constant for a chemical reaction at two different temperatures is given by (A) Kp2/Kp1 = - (∆H/R) (1/T2- 1/T1) (B) Kp2/Kp1 = (∆H/R) (1/T2- 1/T1) (C) Kp2/Kp1 = ∆H (1/T2- 1/T1) (D) Kp2/Kp1 = - (1/R) (1/T2- 1/T1)

Last Answer : (A) Kp2/Kp1 = - (∆H/R) (1/T2- 1/T1)

The efficiency of a Carnot heat engine operating between absolute temperatures T1 and T2(when, T1 > T2) is given by (T1- T2)/T1. The co efficient of performance (CO.P.) of a Carnot heat pump operating between T1 and T2is given by (A) T1/(T1-T2) (B) T2/(T1-T2) (C) T1/T2 (D) T2/T1

Description : The efficiency of a Carnot heat engine operating between absolute temperatures T1 and T2(when, T1 > T2) is given by (T1- T2)/T1. The co efficient of performance (CO.P.) of a Carnot heat pump operating between T1 and T2is given by (A) T1/(T1-T2) (B) T2/(T1-T2) (C) T1/T2 (D) T2/T1

Last Answer : (B) T2/(T1-T2)