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.

explanation: it gains negative electrons, therefore becoming negatively charged. as charge cannot be created, nor destroyed, only transferred, this leaves the wool positive. now that they have opposite charges, the two items attract.

2b2t

explanation:

2b2t