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.**A cylinder of gas is assumed to contain 11.2 kg of butane (C

_{4}H

_{10}). If a normal family needs 20000 kJ of energy per day, the cylinder will last ( Given that Î”H for combustion of butane is −2658 kJ )

Solution:-

Cylinder contains 11.2kg or 193.10 mole butane. (∵molecular mass of butane = 58)

∵ Energy released by 1 mole of butane = −2658

∵ Energy released by 193.10 mole of butane = −2658 × 193.10

= 5.13 × 10

∴ 5.13×10

Cylinder contains 11.2kg or 193.10 mole butane. (∵molecular mass of butane = 58)

∵ Energy released by 1 mole of butane = −2658

∵ Energy released by 193.10 mole of butane = −2658 × 193.10

= 5.13 × 10

^{5}kJ∴ 5.13×10

^{5}÷ 20000 = 25.66 or 26 days**Q2.**Heat of combustion of CH

_{4}, C

_{2}H

_{4}, C

_{2}H

_{6}are -890, -1411 and -1560 kJ/mol respectively. Which has the lowest calorific fuel value in kJ/g ?

Solution:-

calorific value = Heat of combustion per g of fuel,

i.e. for C

calorific value = Heat of combustion per g of fuel,

i.e. for C

_{2}H_{4}, it is −1411÷28, the lowest value.**Q3.**16 g oxygen gas expands at STP to occupy double of its original volume. The work done during the process is :

Solution:-

At STP, 16g O

(V

Now, W = P × (V

= (1×11.2×2)÷0.0821 = 272.84 kcal

At STP, 16g O

_{2}or 1/2 mole O_{2}will occupy 11.2 litre. Thus if volume is doubled, it means(V

_{2}− V_{1}) = 22.4−11.2 = 11.2 litreNow, W = P × (V

_{2}− V_{1}) = 1×11.2 litre atm= (1×11.2×2)÷0.0821 = 272.84 kcal

**Q4.**When two moles of hydrogen expands isothermally against a constant pressure of 1 atm, at 25 °C from 15L to 50L, the work done (in litre atm) will be :

Solution:-

p = 1 atm

Î”V=(50−15)=35L

∴ W = −p.Î”V = −1×35 = −35 atm L

Hence work done by the system on the surroundings is equal to 35 L-atm

p = 1 atm

Î”V=(50−15)=35L

∴ W = −p.Î”V = −1×35 = −35 atm L

Hence work done by the system on the surroundings is equal to 35 L-atm

**Q5.**The enthalpy changes of formation of the gaseous oxides of nitrogen(N

_{2}O and NO) are positive because of :

Solution:-

Due to high bond energy of N≡N, more heat is absorbed to breakup N

Due to high bond energy of N≡N, more heat is absorbed to breakup N

_{2}molecule.**Q6.**The enthalpy change is negative for :

Solution:-

Rest all are endothrmic process. Hydration is always exothermic.

Rest all are endothrmic process. Hydration is always exothermic.

**Q7.**A carnot engine operates between temperature T and 400K (T>400K). If efficiency of engine is 25%, the temperature T is :

Solution:-

Î·

∴ T=533.3 K

Î·

_{efficiency}=(T_{2}− T_{1})/T_{2}, here T_{2}= T and T_{1}= 400K∴ T=533.3 K

**Q8.**A reaction occur spontaneously if :

Solution:-

For spontaneous reaction Î”G = −ve.

Î”G = Î”H − TÎ”S

Î”H = +ve, Î”S = +ve and TÎ”S > Î”H

For spontaneous reaction Î”G = −ve.

Î”G = Î”H − TÎ”S

Î”H = +ve, Î”S = +ve and TÎ”S > Î”H

**Q9.**The resultant heat change in a reaction is same whether it take place in one or several stages. This statement is called

Solution:-

Hess's law

Hess's law

**Q10.**The heat of combustion of ethanol determined by a bomb calorimeter is −670.48 kcal/mol at 25°C. What is Î”U at 25°C for the following reaction ?

C

_{2}H

_{5}OH(l) + 7/2 O

_{2}(g) → 2CO

_{2}(g) + 3H

_{2}O(g)

Solution:-

Experimental determination of heat of reaction by bomb calorimeter represents its value at constant volume i.e. Î”U.

Experimental determination of heat of reaction by bomb calorimeter represents its value at constant volume i.e. Î”U.