Work Done in Reversible Adiabatic Process
Trending Questions
Q.
In a typical combustion engine, the work done by a gas molecule is given by
where is the displacement, is the Boltzmann constant and is the temperature. If and are constants, dimensions of will be
Q.
What is an isothermal process?
Q. How the change in internal energy is zero in the isothermal process? Explain?
Q. Match the Column - I with Column - II
Column−I(IdealGas)Column - II (Related equation)(A) Reversible isothermal process(P)W=2.303nRTlog(P2/P1)(B) Reversible adiabatic process(Q)W=nCv, m(T2−T1)(C) Irreversible adiabatic process(R)PV=nRT(D) Irreversible isothermal process(S)W=−∫vfviPext.dV
Column−I(IdealGas)Column - II (Related equation)(A) Reversible isothermal process(P)W=2.303nRTlog(P2/P1)(B) Reversible adiabatic process(Q)W=nCv, m(T2−T1)(C) Irreversible adiabatic process(R)PV=nRT(D) Irreversible isothermal process(S)W=−∫vfviPext.dV
- A) P, R, S
B) Q, R, S
C) Q, R, S
D) R, S - A) P, S
B) Q, S
C) Q, R, S
D) R, S - A) P
B) Q, S
C) Q, S
D) R, S - A) P, Q, R, S
B) Q, S
C) Q, S
D) R, S
Q. For the process to occur under adiabatic conditions, the correct condition is:
- △T=0
- △P=0
- w=0
- △q=0
Q.
An ideal gas at has compressed adiabatically to of its original volume. If then the rise in temperature will be
Q. Calculate the work done when 10 moles of ideal gas expand isothermally and reversibly from a pressure of 10 atm to 1 atm at 300 K.
(Given : R=2 cal K−1 mol−1)
(Given : R=2 cal K−1 mol−1)
- −13.82 kcal
- 13.82 kcal
- −6 kcal
- 6 kcal
Q. Match the Column - I with Column - II
Column−I(IdealGas)Column - II (Related equation)(A) Reversible isothermal process(P)W=−2.303nRTlog(V2/V1)(B) Reversible adiabatic process(Q)W=nCv, m(T2−T1)(C) Irreversible adiabatic process(R)PV=nRT(D) Irreversible isothermal process(S)W=−∫vfviPext.dV
Column−I(IdealGas)Column - II (Related equation)(A) Reversible isothermal process(P)W=−2.303nRTlog(V2/V1)(B) Reversible adiabatic process(Q)W=nCv, m(T2−T1)(C) Irreversible adiabatic process(R)PV=nRT(D) Irreversible isothermal process(S)W=−∫vfviPext.dV
- A) P, R, S
B) Q, R, S
C) Q, R, S
D) R, S - A) P, S
B) Q, S
C) Q, R, S
D) R, S - A) P
B) Q, S
C) Q, S
D) R, S - A) P, Q, R, S
B) Q, S
C) Q, S
D) R, S
Q. One mole ideal monoatomic gas is heated according to path AB and AC. If temperature of state B and C are equal. Calculate qACqAB×10
Q. An ideal monoatomic gas undergoes adiabatic free expansion from 16 bar, 2 L, 300 K to 16 L.
- The final pressure of gas becomes zero
- The final pressure of gas becomes 2 bar
- The final temperature of gas becomes 300×(216)23 K
- The final temperature of gas becomes 300 K
Q. One gram mol of a diatomic gas (γ=1.4) is compressed adiabatically so that its temperature rises from 27oC to 127oC. The work done will be
- 207.85 joules
- 2078.5 joules
- 207.85 ergs
- None of the above
Q. The reversible expansion of an ideal gas under adiabatic and isothermal condition is shown in the figure. Which of the following statements is incorrect?
- T1=T2
- T3>T1
- Wisothermal>Wadiabatic
- △Uisothermal>△Uadiabatic
Q. A system works under cyclic process as follows
Heat absorbed during the process is
Heat absorbed during the process is
- 227×102 J
- 227×103 J
- 227×104 J
- 227×105 J
Q. One mole of an ideal gas expands isothermally from initial state ( 1 atm, 22.4 L, 273 K ) to the final state (0.25 atm, 89.6 L) in reversible manner. One mole of the same gas with the same initial state in a different cylinder expands against a constant external pressure of 0.25 atm to final volume 89.6 L. The difference in work done in J is
(Take 1 L atm=101 J, R=8.3 J K−1 mol−1, log2=0.3)
(Take 1 L atm=101 J, R=8.3 J K−1 mol−1, log2=0.3)
Q. A gas expands adiabatically at constant pressure such that T∝V−12
The value of γ(Cp, m/Cv, m) of the gas will be:
The value of γ(Cp, m/Cv, m) of the gas will be:
- 1.70
- 1.30
- 2
- 1.50
Q. One mole of a gas is heated at constant pressure to raise its temperature by 1∘C The work done in joules is
- −4.3
- −8.314
- −16.62
- Unpredictable
Q. The rate constants of a reaction at 500 K and 700 K are 0.02s−1and0.07s−1 respectively. Calculate the values of Ea and A.
Q.
Which of the following is true in case of reversible adiabatic expansion?
Q. Which of the following statement (s) is / are correct?
- For adiabatic expansion of an ideal gas, TVγ−1 = constant
- Work done in reversible isothermal expansion is greater than that done in reversible adiabatic expansion for the same increase of volume
- Buffer capacity is maximum when concentration of weak acid and salt of its conjugate base is equal
- Equilibrium constant of an exothermic reaction decreases with increase of temperature generally
Q. When one mole of monoatomic ideal gas at temperature T undergoes adiabatic change reversibly, change in volume is from 1 L to 2L, the final temperature in Kelvin would be
- T22/3
- T+23×0.0821
- T
- T−23×0.0821
Q. One mole of an ideal gas expands isothermally from initial state ( 1 atm, 22.4 L, 273 K ) to the final state (0.25 atm, 89.6 L) in reversible manner. One mole of the same gas with the same initial state in a different cylinder expands against a constant external pressure of 0.25 atm to final volume 89.6 L. The difference in work done in J is (Round the answer to nearest integer)
(Take 1 L atm=101 J, R=8.3 J K−1 mol−1, log2=0.3)
(Take 1 L atm=101 J, R=8.3 J K−1 mol−1, log2=0.3)
Q. An ideal gas whose adiabatic exponent γ is expanded according to the law P=αV, where α is a constant. The initial volume of the gas is equal to V0. As a result of expansion, the volume increases 4 times. Select the correct option for above information.
- Molar heat capacity of gas in the process is R(γ+1)2(γ−1)
- Work done by the gas is − 15V20α2
- Molar heat capacity of gas in the process is R(γ−1)2(γ+1)
- Work done by the gas is + 15V20α2
Q. 5 moles of an ideal gas (Cv=52R) was compressed adiabatically against a constant pressure of 5 atm , which was initially at 250 K and 1 atm pressure. The work done in the process is equal to:
- 1450 R
- 3562 R
- 2500 R
- 5000 R
Q. Gas equation pV = nRT is obey by ideal gas in (adiabatic/isothermal/both)?
Q. A gas (Cv, m=52R) behaving ideally was allowed to expand reversibly and adiabatically from 1 litre to 32 litre. Its initial temperature was 227oC. The molar enthalpy change (in J/mol) for the process is −x×R. Find the value of x
Q. Calculate pH of a resultant solution of 25 mL of 0.1 M HCl, 50 mL of 0.01 M HNO3 and 25 mL of 0.1 M NaOH.
- 5.05
- 4.15
- 4.75
- 2.3
Q. Work Done in reversible adiabatic process is given by:
- nR(γ−1)(T2−T1)
- 2.303 RT logV1V2
- None of these
- 2.303 RT logV2V1
Q. Derive an expression of the mechanical work done in an isothermal reversible expansion of an ideal gas.
Q. One mole of an ideal gas at 300K in thermal contact with surroundings expands isothermally from 1.0L to 2.0L against a constant pressure of 3.0atm.
In this process, the change in entropy of surroundings (ΔSsurr) in JK−1 is (1L atm =103J):
- +5.763
- +1.013
- −1.013
- −5.763
Q. For which process does ΔU=0 holds true for ideal gas:
- Adiabatic process
- Isochoric process
- Cyclic process
- Isothermal process