Heat Capacity at Constant Volume
Trending Questions
Q. Two moles of an ideal gas is heated at constant pressure of one atmosphere from 27oC to 127oC. If Cv, m=20+10−2T JK−1 mol−1, then q and △U for the process are respectively:
- 6362.8 J, 4700 J
- 3037.2 J, 4700 J
- 3181.4 J, 2350 J
- 7062.8, 5400 J
Q. At constant volume, the specific heat of a gas is 0.075 cal K−1 g−1 and its molecular weight is 40 g mol−1. The atomicity of the gas is:
Given: R = 2 cal K−1mol−1
Given: R = 2 cal K−1mol−1
- Monoatomic
- Diatomic
- Triatomic
- None of the above
Q. The standard free energy change (ΔG∘) for 50% dissociation of N2O4 intoNO2 at 27∘C and 1 atm pressure is −x J mol−1. The value of x is (Nearest Integer)
[Given:R=8.31 J K–1mol–1, log(1.33)=0.1239, ln(10)=2.3]
[Given:R=8.31 J K–1mol–1, log(1.33)=0.1239, ln(10)=2.3]
Q. What is the value of the change in internal energy at 1 atm in the following process?
H2O(l, 323 K)→H2O(g, 423 K)
Given : Cv, m(H2O, 1)=75.0 JK−1;Cp, m(H2O, g)=33.314JK−1mol−1;ΔHvapat373K=40.7kJ/mol
H2O(l, 323 K)→H2O(g, 423 K)
Given : Cv, m(H2O, 1)=75.0 JK−1;Cp, m(H2O, g)=33.314JK−1mol−1;ΔHvapat373K=40.7kJ/mol
- 42.910 kJ/mol
- 43.086 kJ/mol
- 42.600 kJ/mol
- 49.600 kJ/mol
Q. Temperature of one mole of helium gas is increased by 4∘. Thus, increase in internal energy is
- 6 cal
- 12 cal
- 3 cal
- 24 cal
Q. If Cv=4.96 cal/mol.K then increase in internal energy when temperature of 2 moles of this gas is increased from 340 K to 342 K
- 19.84 cal
- 27.80 cal
- 13.90 cal
- 9.92 cal
Q. Choose the correct option related to the specific heat at constant volume Cv :
- All of these
- Cv=Rγ−1
- Cv=Rγ
- Cv=12γR
Q. For a gas (R/Cv)=0.67 the gas is made up of molecules which are
(R = gas constant and Cv = heat capacity at constant volume)
(R = gas constant and Cv = heat capacity at constant volume)
- monoatomic
- diatomic
- polyatomic
- mixture of gases
Q.
In adiabatic irreversible expansion of an ideal gas against a constant external pressure P2, the P - V work is given by ?
−Wirr=nR(T2−T1P2P1)
−Wirr=nR(T2+T1P2P1)
−Wirr=nRT1(1−P2P1)
−Wirr=nRT2(1−P2P1)
Q. A thermally insulated vessel is divided into two compartments A and B by a partition of insulating material. Compartment A has 0.1 mole of Helium at 427∘C and 1 atm and compartment B has x mol of Neon at 127∘C and 1 atm. If partition is removed and final temperature of gas is 227∘C, then the value of 10x is . [Assume Cv=3R2 for both gases ]
Q. A heat engine carries 1 mole of an ideal monoatomic gas around the cycle as shown in the figure, the amount of heat added in the process AB and heat removed in th process CA are:
- qAB=450R and qCA=−225 R
- qAB=450R and qCA=−450 R
- qAB=375R and qCA=−450 R
- qAB=450R and qCA=−375 R
Q. What is the final temperature of 0.10 mole monoatomic ideal gas that performs 75 cal of work adiabatically if the initial temperature is 227∘C?
(use R=2 cal K−1 mol−1)
(use R=2 cal K−1 mol−1)
- 250 K
- 300 K
- 350 K
- 750 K
Q.
In adiabatic irreversible expansion of an ideal gas against a constant external pressure P2, the P - V work is given by ?
−Wirr=nR(T2−T1P2P1)
- −Wirr=nR(T2+T1P2P1)
−Wirr=nRT1(1−P2P1)
−Wirr=nRT2(1−P2P1)
Q. If K=1.19×10-4 at 623K. Find 🔼G
Q. For the reaction, 2NO(g)+O2(g)→2NO2(g) the value of ΔH is −113.1 kJ. If 6.0 moles of NO reacts with 3.0 moles of O2 at 1.0 atm and 300 K to form NO2. Calculate the work done against a pressure of 1.0 atm.
(Take R=112 L atm mol−1K−1)
(Take R=112 L atm mol−1K−1)
- 7.56 kJ
- −75 J
- −7.56 kJ
- 75 J
Q. A solution containing 28 g phosphorous in 315 g CS2 (b.pt. 46.3oC) boils at 47.98oC, Kb for CS2 is 2.34 K mol−1kg. Its molecular formula is Px. Find x. Assume its complete association.
Q. For reaction, 2NOCl(g)⇌2NO(g)+Cl2(g);Kc at 427oC is 3×10−6Lmol−1. The value of Kp is nearly:
- 7.50×10−5
- 2.50×10−5
- 2.50×10−4
- 1.75×10−4
Q. When 10g of Al is used for reduction in each of the following alumino thermic reactions, which reaction would generate more heat and by how much ?
a. 2Al+Cr2O3→Al2O3+2Cr
b. 2Al+Fe2O3→Al2O3+2Fe
a. 2Al+Cr2O3→Al2O3+2Cr
b. 2Al+Fe2O3→Al2O3+2Fe
Standard heat of formation of Al2O3, Cr2O3 and Fe2O3 are −1676kJ, −1141kJ and −822.2kJ, respectively.
- Reaction (b) produces more heat by 59.04kJ
- Reaction (a) produces more heat by 59.04kJ
- Reaction(a) and (b) produces equal heat
- None of these
Q. A 4:1 molar mixture of He and CH4 is contained in a vessel at 20 bar pressure. Due to a hole in the vessel, the gas mixture leaks out. What is the composition ratio of the mixture effusing out initially?
- rHerCH4=1:4.
- rHerCH4=8:1.
- rHerCH4=1:8.
- None of these
Q.
ΔGo for the reaction:
AgCl(s)+1/2H2(g)⟶Ag(s)+H++Cl− is −21.52kJΔGo for the reaction:
2AgCl(s)+H2(g)→2Ag(s)+2H++2Cl− is:
- −21.52KJ
- −10.76KJ
- −45.04KJ
- −43.04KJ
Q. Reaction of methanol with dioxygen was carried out and ΔU was found to be −726 kJmol−1 at 298K. The enthalpy change for the reaction will be
CH3OH(l)+32O2(g)→CO2(g)+2H2O(l); ΔU=−726 kJmol−1
CH3OH(l)+32O2(g)→CO2(g)+2H2O(l); ΔU=−726 kJmol−1
- −727kJmol−1
- +741.5kJmol−1
- +727.2kJmol−1
- −741.5kJmol−1
Q. A new flurocarbon of molar mass 102gmol1 was placed in an electrically heated vessel. When the pressure was 650 torr, the liquid boiled at 77∘C. After the boiling point had been reached, it was found that a current of 0.25 A from a 12.0 volt supply passed for 600 sec vaporises 1.8 g of the sample. The molar enthalpy & internal energy of vaporisation of new flourocarbon will be :
- ΔH=102kJ/mol, ΔE=99.1kJ/mol
- ΔH=95kJ/mol, ΔE=100.3kJ/mol
- ΔH=107kJ/mol, ΔE=105.1kJ/mol
- ΔH=92.7kJ/mol, ΔE=97.4kJ/mol
Q. One kilogram water at 0oC is brought into contact with a heat reservoir at 100oC. Find:
(a) Change in entropy when temperature reaches to 100oC.
(b) What is the change in entropy of reservoir?
(c) Change in entropy universe.
(d) The nature of process.
(a) Change in entropy when temperature reaches to 100oC.
(b) What is the change in entropy of reservoir?
(c) Change in entropy universe.
(d) The nature of process.
Q. For the reaction Ag+(aq)+Cl−(aq)⇌AgCl(s); the ΔGo values for Ag+(aq), Cl−(aq) and AgCl(s) are + 77, -129 and -109 kJ mol−1. Write the cell representation of above reaction and calculate Ksp of AgCl at 298 K.
(Multiply the ans by 1010)
(Multiply the ans by 1010)
Q. Two moles of an ideal gas is heated at constant pressure of one atmosphere from 27oC to 127oC. If Cv, m=20+10−2T JK−1 mol−1, then q and △U for the process are respectively:
- 6362.8 J, 4700 J
- 3037.2 J, 4700 J
- 7062.8, 5400 J
- 3181.4 J, 2350 J
Q.
CH3−O||C−CH3(g)⇌CH3−CH3(g)+CO(g)
The initial pressure of CH3COCH3 is 100 mm of Hg. When equilibrium is achieved, the mole fraction of CO(g) is 1/3 hence, Kp is:- 100 mm
- 50 mm
- 25 mm
- 150 mm
Q. If, Kp=Kc×[RT]Δn, for the reaction ; 2CO2(g)+2CaCO3(s)⇌2CaC2(s)+5O2(g); Δn is equal to :
Q. If Cv=4.96 cal/mol.K then increase in internal energy when temperature of 2 moles of this gas is increased from 340 K to 342 K
- 27.80 cal
- 19.84 cal
- 13.90 cal
- 9.92 cal
Q. The molar heat capacity of oxygen gas is given by the expression Cv=a+bT+cT2 where a, b and c are constants. What will be the change in the internal energy of 8 g of oxygen, if it is heated from 200 K to 300 K at constant volume? Assume oxygen to be an ideal gas. (Given, a=1.2 JK−1mol−1, b=12.8×10−2 JK−2mol−1, c=1.0×10−7 JK−3mol−1).
- 850.60 J
- 950.50 J
- 830.16 J
- 853.62 J
Q. ΔGo of the cell reaction AgCl(s)+1/2H2(g)→Ag(s)+H+(aq.)+Cl−(aq) is −21.52kJ.
Calculate the EMF for the cell reaction:
2AgCl(c)+H2(g)→Ag(s)+2H+(aq.)+2Cl−(aq.)
2AgCl(c)+H2(g)→Ag(s)+2H+(aq.)+2Cl−(aq.)
- 0.223V
- 0.112V
- 0.446V
- 0.337V