Aparticular container holds 4.67 mol of neon gas. the volume of this container can be altered by sliding a piston in or out. the volume is changed from 8.90 l to 7.60 l while at the same time the temperature is changed from 344 k to 294 k. the molar heat capacity, cv, m, for neon is 12.47 j/(mol · k). assume that this value will not change over the given temperature range. what is the change in entropy for the gas?

15.24 J/K

Explanation:

First, let's write the expression to calculate the change in entropy:

ΔS = n*C*ln(T2/T1)

C is the heat capacity. It will be Cp if the heating is isobaric, and Cv if it is isochoric. (In this case is Cv)

Now, in order to do this, we need to calculate first the pressure of the gas at first, and then, after the chance:

P = nRT/V

P1 = 4.67 *¨0.082 * 344 / 8.9 = 14.801 atm

P2 = 4.67 * 0.082 * 294 / 7.6 = 14.814 atm

Now, let's calculate the change in entropy:

ΔS = 4.67 * (12.47 + 8.314) ln(344/294)

ΔS  = 15.24 J/K

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"