Thermodynamics is a very important branch of both chemistry and physics. It deals with the study of energy, the conversion of energy between different forms and the ability of energy to do work. Thermodynamics is an important chapter for JEE Mains, JEE Advance NEET and many others exams. As this chapter is present in both chemistry and physics and there is only a minor difference between them hence it becomes more important topic. It is not a tough topic, it only requires practice. This topic has been given good weightage in all engineering exams. So don't skip this and prepare well. All the best !.
Q1. Thermochemistry is the study of relationship between heat energy and :
Solution:-
The branch deals with interconversion of heat and chemical energy.
The branch deals with interconversion of heat and chemical energy.
Q2. If gas, at constant temperature and pressure expands then its
Solution:-
Internal energy of a gas depends upon its pressure and temperature. Thus, if a gas expands at constant temperature and pressure, then its internal energy remains same.
Internal energy of a gas depends upon its pressure and temperature. Thus, if a gas expands at constant temperature and pressure, then its internal energy remains same.
Q3. The total amount of energy in the universe is fixed, but:
Solution:-
Entropy (a measure of disorder) of universe is increasing towards maximum. This is second law of thermodynamics.
Entropy (a measure of disorder) of universe is increasing towards maximum. This is second law of thermodynamics.
Q4. ΔE° of combustion of isobutylene is −x kJ/mol. The value of ΔH° is :
Solution:-
(CH3)2C=CH2(g) + 6O2(g) → 4CO2(g) + 4H2O(l)
Δng = 4−7 = −3
We know that ΔH = ΔE + ΔngRT
= ΔE − (Δm)RT (∵ Δng = −ve)
∴ ΔH < ΔE
(CH3)2C=CH2(g) + 6O2(g) → 4CO2(g) + 4H2O(l)
Δng = 4−7 = −3
We know that ΔH = ΔE + ΔngRT
= ΔE − (Δm)RT (∵ Δng = −ve)
∴ ΔH < ΔE
Q5. The work done during the process when 1 mole of gas is allowed to expand freely into vacuum is:
Solution:-
W = −Pext.(V2−V1)
∵ Pext. = 0
∴ W = 0
W = −Pext.(V2−V1)
∵ Pext. = 0
∴ W = 0
Q6. The heat of combustion of rhombic and monoclinic sulphur are 70.96 and 71.03 kcal. The heat of transition of SR→M is:
Solution:-
SR + O2 → SO2; ΔH = −70.96 kcal.....(i)
SM + O2 → SO2; ΔH = −71.03 kcal.....(ii)
By eq. (i) − (ii)
SR → SM; ΔH = 0.07 kcal or 70 cal
SR + O2 → SO2; ΔH = −70.96 kcal.....(i)
SM + O2 → SO2; ΔH = −71.03 kcal.....(ii)
By eq. (i) − (ii)
SR → SM; ΔH = 0.07 kcal or 70 cal
Q7. From the following bond energies :
H−H bond energy : 431.37 kJ/mol
C=C bond energy : 606.10 kJ/mol
C−C bond energy : 336.49 kJ/mol
C−H bond energy : 410.50 kJ/mol
Enthalpy for the reaction,
H−H bond energy : 431.37 kJ/mol
C=C bond energy : 606.10 kJ/mol
C−C bond energy : 336.49 kJ/mol
C−H bond energy : 410.50 kJ/mol
Enthalpy for the reaction,
Solution:-
Q8. 1 mole of gas occupying 3 L volume is expanded against a constant external pressure of 1 atm to a volume of 15 litre. The work done by the system is:
Solution:-
−Wirr. = Pext.(V2−V1)
= 1 × (15−3) = 12 litre atm
= (12 × 1.987 × 4.184) / 0.0821 = 1.215 × 103 J
−Wirr. = Pext.(V2−V1)
= 1 × (15−3) = 12 litre atm
= (12 × 1.987 × 4.184) / 0.0821 = 1.215 × 103 J
Q9. Which one of the following is not a state function?
Solution:-
Macroscopic properties which determine the state of a system are referred as state functions. The change in the state properties depends only upon the initial and final state of the system. All thermodynamic functions are state functions except work and heat.
Macroscopic properties which determine the state of a system are referred as state functions. The change in the state properties depends only upon the initial and final state of the system. All thermodynamic functions are state functions except work and heat.
Q10. A closed flask contains water in all its three states, solids, liquid and vapour at 0°C. In this situation the average KE of the water molecule will be:
Solution:-
KE = (3/2)RT
KE = (3/2)RT