What is the change in molar entropy of helium in the following process ? 1 He (298 K, 1…

Chemistry · JEE Advanced · NTA Exams — Chemical Thermodynamics

1. 41.8 J K^–1 mol^–1
2. 4.18 J K^–1 mol^–1
   
3. – 41.8 J K^–1 mol^–1
4. – 4.18 J K^–1 mol^–1

Answer

(C) – 41.8 J K –1 mol –1

Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.

Related practice questions

• The enthalpy of neutralization of a strong acid by a strong base is –57.32 kJ mol –1 . The enthalpy of…
• An ideal gas undergoes a four step cycle as shown in the P - V diagram. During this cycle, heat is absorbed…
• Spontaneous adsorption of a gas on solid surface is an exothermic process because (2004)
• The correct option(s) about entropy ( S ) is(are) [ R = gas constant, F = Faraday constant, T = Temperature]
• 70 calorie of heat are required to raise the temperature of 2 mole of an ideal gas at constant pressure from…
• Heat released in neutralization of strong acid and strong base is 13.4 kcal/mol. The heat released on…
• ∆H vap = 30 kJ/mol and ∆S vap = 75 Jmol –1 K –1 . Find temperature of vapour, at one atmosphere. (2004)
• Heat of neutralization of a strong dibasic acid in dilute solution by NaOH is nearly

More Chemical Thermodynamics questions · Browse all practice questions