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NEET2010ChemistryThermodynamics (C)Actual

For vaporisation of water at 1 ~atm pressure, the values of H and S are 40.63 ~kJ ~mol ⁻¹ and 108.8 JK ⁻¹ ~mol ⁻¹ , respectively. The temperature when Gibbs energy change ( G) for this transformation will be zero, is

Options

  1. A273.4 ~K
  2. B393.4 ~K
  3. C373.4 ~K
  4. D293.4 ~K

Correct answer

C. 373.4 ~K

Step-by-step solution

aligned G & = H - T S G & =0( given ) H & = T S , T & = 40.63 10^3 108.8 =373.4 ~K aligned

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