A sample of helium (He) gas initially at 19°C and 1.0 atm is expanded from 1.4 L to 2.8 L and simultaneously heated to 35°C. Calculate the entropy change for the process.

Explanation:

The given data is as follows.

  = (19 + 273) K = 292 K,   = 1.0 atm,

  = 1.4 L

Therefore, moles of gas present will be calculated as follows.

          n =  

             =

             =

             = 0.058 mol

Since, neon is a monoatomic ideal gas hence, expression for its constant volume heat capacity is as follows.

As the gas is expanding to a volume () of 2.8 L and it is heated to or (35 + 273) K = 308 K.

So, entropy change for this reaction will be calculated as follows.

               =

            =

            =

            =

            = 0.372 J/K

Thus, we can conclude that the entropy change for the process is 0.372 J/K.

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