Question

For isothermal process AE = 0 and w = -2.303 nRT log10 P2
lustration 10. At 27°C, one mole of an ideal gas compressed isothermally and reversibly from a pressure
of 2 atm to 10 atm. Calculate internal energy and q in calorie.
PA
Solution
2
w = -2.303 x 1 x 2 x 300 x log
10
w = + 2.303 x 600 x log 5
w = + 2.303 x 600 x 0.699
+ 965.87 Cal
W =
For isothermal process
W = -
..q = - 965.87 cal
this is the complete solution given in my text I just want to know why is the value of R used as 2 cal /mol k and not as 0.0821 l-atm /mol k .. I know the answer has been asked in calorie but in the question the pressure is given in atm how can we consider it as calorie to convert l-atm to calorie we have to divide it by 24 right
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Answer 1

Answer:

Ellison, A.J.G.; Navrotsky, A.

1992-01-01

Using high-temperature solution calorimetry in molt 2PbO · B 2 O 3 , the enthalpy of reaction of the formation of zircon, ZrSiO 4 , from its constituent oxides has been determined: Δ 4 H 977 (ZrSiO 4 ) = -27.9 (±1.9) kJ/mol. With previously reported data for the heat contents of ZrO 2 SiO 2 and ZrSiO 4 and standard-state enthalpies of formation of ZrO 2 and SiO 2 , we obtain Δ f H 298 degrees. (ZrSiO 4 ) = -2034.2 (±3.1) kJ/mol and Δ t G 298 degrees (ZrSiO 4 ) = -1919.8 kJ/mol. The free energy value is in excellent agreement with a range previously estimated from solid-state reaction equilibria. At higher temperature also the data are in close agreement with existing data, though the data sets diverge somewhat with increasing T. In this paper the limitations of the data for predicting the breakdown temperature of zircon into its constituent oxides are discussed