Thermodynamics ...

Direction (Q. Nos. 1-10) This section contains 10 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.

Q. Entropy change for the following reversible process is 1 mole H₂O

(l, 1 atm, 100°C ) 1 mole H₂O (g , 1 atm, 100°C)(ΔH_vap = 40850 J mol^-1)

Entropy change when 2 moles of an ideal gas expands reversibly from an initial volume of 1 dm³ to a final volume of 10 dm³ at a constant temperature of 298 K is

3 moles of a diatomic gas are heated from 127° C to 727° C at a constant pressure of 1 atm. Entropy change is (log 2.5 = 0 .4)

10 dm³ of an ideal monoatomic gas at 27° C and 1.01 x 10⁵ Nm^-2 pressure are heated at constant pressure to 127°C. Thus entropy change is

Exactly 100 J of heat was transferred reversibly to a block of gold at 25.00° C from a thermal reservoir at 25.01 °C, and then exactly 100 J of heat was absorbed reversibled from the block of gold by a thermal reservoir at 24.99° C. Thus entropy change of the system is

Given

I. C (diamond) + O₂(g) → CO₂(g) ; ΔH° = - 91.0 kcdl mol^-1
II. C(graphite) + O₂(g) → CO₂(g) ; ΔH° = - 94.0 kcal mol^-1

Q. At 298 K, 2.4 kg of carbon (diamond) is converted into graphite form. Thus, entropy change is

Consider a reversible isentropic expansion of 1.0 mole of an ideal monoatomic gas from 25°C to 75°C. If the initial pressure was 1.0 bar, final pressure is 

Consider the following figure representing the increase in entropy of a substance from absolute zero to its gaseous state at some temperature

Q. ΔS° (fusion) and ΔS° (vaporisation) are respectively indicated by 

ΔH_vap = 30 kJ mol^-1 and ΔS_vap = 75 J mol^-1 K^-1. Thus, temperature of the vapour at 1 atm is

[IIT JEE 2004]

For the process, and 1 atmosphere pressure, the correct choice is

[JEE Advanced 2014]