Second law of thermodynamics - class-XI

Given R/Cp = 0.4, we find Cp = R/0.4 = 2.5R. Since Cp - Cv = R, we get Cv = 1.5R. The ratio of specific heats γ = Cp/Cv = 2.5R/1.5R = 5/3 = 1.67. This corresponds to a monatomic gas with f = 3 degrees of freedom. Option B correctly asks for Cv and γ values, which are the natural quantities to calculate from the given.

Efficiency n = 1 - (T2/T1). For n1: T1 = 473K, T2 = 273K, n1 = 1 - (273/473) = 200/473. For n2: T1 = 273K, T2 = 73K, n2 = 1 - (73/273) = 200/273. Ratio n1/n2 = (200/473) * (273/200) = 273/473 approx 0.577.

Efficiency n = W/Q1 = (Q1-Q2)/Q1. Coefficient of performance (COP) for a refrigerator is Q2/W. Since W = Q1 - Q2, COP = Q2/(Q1-Q2). Dividing by Q1 gives (Q2/Q1) / (1 - Q2/Q1). Since n = 1 - Q2/Q1, Q2/Q1 = 1 - n. Substituting gives (1-n)/n = (1/n) - 1.

Heat is defined as the energy which passes between two objects because of the difference in temperature. Thus energy may pass between them, but not heat. It is possible that the energy absorbed may increase the temperature, particularly if work is done, but this is not heat transfer.

The second law of thermodynamics states that the entropy of an isolated system never decreases, because isolated systems spontaneously evolve toward thermodynamic equilibriumthe state of maximum entropy

The second law of thermodynamics says that heat flows 'naturally' from an object at a higher temperature to an object at a lower temperature, and heat doesn’t flow in the opposite direction of its own accord, since the entropy in such a process increases. 

The second law of thermodynamic states that the entropy of an isolated system increases over time and remain constant for reversible process. 

The second law of thermodynamics states , that it is impossible to construct a heat engine , operating in cycle , which extracts heat and can convert it all to useful work .In other words , it is impossible for a heat engine to convert heat completely in work .This statement is given by Kelvin & Plank .

Second law of thermodynamics states that in a cyclic process heat cannot be converted completely into work because attainment of $0K$ temperature is impossible.

The second law of thermodynamics states that the total entropy of an isolated system always increases over time, or remains constant in ideal cases where the system is in a steady state or undergoing a reversible process. Thus an opened bottle allows gas to escape since the entropy increases(disorderliness increases) over time.

The second law of thermodynamics states , that it is impossible to construct a heat engine , operating in cycle , which extracts heat and can convert it all to useful work .In other words , it is impossible for a heat engine to convert heat completely in work .

In a heat engine , if $Q _{1}$ is the heat absorbed from reservoir , $W$ be the work done by engine and let $Q _{2}$ be the heat given to sink , then by the first law of thermodynamics 

The Second Law of Thermodynamics tells us that the total amount of disorder, or entropy, in the universe is increasing. The entropy in a particular system can decrease, as with water molecules when they turn to ice, but only if the entropy in the surroundings of that system increases to an equal or greater extent. The Second Law of Thermodynamics holds, but only because the surroundings are gaining entropy, so the correct answer is E . Answer D refers to the key part of the answer, but gives the wrong information about the change in entropy of the surroundings. 
Answer C is not correct. This is an explanation for why the water does not lose heat when it freezes: it is, in fact, losing internal energy. This is an instance of the First Law of Thermodynamics, which states that the change in a systems internal energy is equal to the value of the heat transfer in the system minus the work done by the system. 

The second law of thermodynamics states that when energy change from one form to another form, entropy in a closed system increases.

second law of thermodynamics states that ,

Entropy is related to the degree of order in the structure of a system and the second law of thermodynamics means that the system always changes from a more ordered state to a less ordered state in any isolated system.

Efficiency of a heat engine is defined as the ratio of net work done per cycle by the working substance ($W$) to the heat absorbed per cycle from the source ($Q _{1}$) ,

First law of thermodynamics just states that $dQ = dW + dU $. 

Work done at constant pressure W = n * R * deltaT. n = 1 mole, R = 1.987 cal/mol*K, deltaT = 25 - 15 = 10 K. W = 1 * 1.987 * 10 = 19.87 cal.

Heat removed = 60 kJ/min. Work input = Heat / COP = 60 / 1.2 = 50 kJ/min. Total time = 30 days * 4 hours/day * 60 min/hour = 7200 min. Total work = 50 kJ/min * 7200 min = 360,000 kJ. 1 unit = 1 kWh = 3600 kJ. Units = 360,000 / 3600 = 100 units. Cost = 100 units * 6 Rs/unit = 600 Rs.

Critical temperature is the temperature above which a gas cannot be liquefied, regardless of the pressure applied.

The given statement is false.

The heat energy released during the freezing of water goes to the surrounding which increases the entropy of the surrounding. Thus although the entropy of the water decreases, the entropy of the surrounding increases such that the net entropy of system (water) and surrounding is greater than zero which does not violate the Second Law.

An inventor claims to have developed an engine that takes in 1000 J of heat and produces 1500 J of work during each cycle.The proposed engine claims to produce more work in a cyclic process than the amount of heat that is supplied, so it is in violation of the first law of thermodynamics.

$L=80 \dfrac{cal}{g}$

Given, $Q _h=700 J$ and $Q _c=500 J$

Given, the temperature of hot reservoir, $T _h=900 K$ and temperature of cold reservoir, $T _c=300 K$

The boiling point of water is $100 ^oC$ and freezing point of water is $0^o C$

$\Delta Q=\Delta U+\Delta W\ \Delta Q-\Delta W=\Delta U$