During the test of an internal combustion engine, 3.00 L of nitrogen gas at 18.5 °C was compressed suddenly (and irrevers- ibly) to 0.500 L by driving in a piston. In the process, the tempera- ture of the gas increased to 28.1°C. Assume ideal behavior, what is the change in entropy of the gas?

Answer : The change in molar entropy of the sample is -14.22 J/K.mol

Explanation :

To calculate the change in molar entropy we use the formula:

where,

= change in molar entropy

n = number of moles = 1 mole

= final temperature =

= initial temperature =

= final volume = 0.500 L

= initial volume = 3.00 L

= heat capacity diatomic gas =

R = gas constant = 8.314 J/mol.K

Now put all the given values in the above formula, we get:

Therefore, the change in molar entropy of the sample is -14.22 J/K.mol

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

c. 4.54 x 10–12 j happens to be correct