II. Practice An ideal gas occupies 5 L at atmospheric pressure and 300 K (point A). It is warmed at constant volume to 3 atm (point B). Then it is allowed to expand isothermally to 1 atm (point C) and at last compressed isobarically to its original state. A. How many moles of gas are being used? B. Find the temperature at point C. C. Find the work done on the gas in each process. D. Find the amount of heat added to/removed from the gas in one cycle.

The process can be represented as shown in the figure below; having got the diagram, we can solve for the questions.

A. the number of moles of gas used

n = PV/ RT = (1.013 *10^5 Pa) * (5.0 *10^-3 m^3) / (8.314 * 300)

n =  5.065 * 10^2 / 2494.2

n = 0.00203 *10^2

n = 0.203 moles

B. Temperature at point C (Tc)

Pa/Ta= Pb/Tb

Tb = Pb *Ta / Pa

Tb = 3 * 300 / 1

Tb = 900 K

Since Tb = Tc = 900 K

C. For process AB,

work done is zero

For process BC,

work done = -nRTbln (Vc/Vb)

W = -(0.203 * 8.314 * 900 ln (3)  

W = -(1.518 kJ ln 3

W = -1.67 kJ

For process CA,

W = -P V =-nRT

W = -(0.203 * 8.314 * (-600))

W = 1.01 kJ

Explanation:

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

it's all four options.