Question

When 2.50 moles of N₂(g) is heated in a piston having a constant pressure of 1.75 atm from 0.00 ^oC to 125 ^oC, calculate the work, expressed in J, associated with this process. Interpret the sign of your answer.

Answer 1

Ans. Given,

            Number of moles = 2.50 mol

            Initial temperature = 0.00⁰C = 273.15 K

            Final temperature = 125.0⁰C = 398.15 K

            Change in temperature, dT = 398.15 K - 273.15 K = 125.0 K

# Since volume of a gas is proportional to temperature, change in temperature is equal to the change in volume.

So,

Increase in volume during expansion can be calculated using ideal gas equation as follow-

Using Ideal gas equation for change in V and T:

P (dV) = n R (dT)    - equation 1

            Where, P = pressure in atm

            dV = change in volume in L                       

            n = number of moles

            R = universal gas constant= 0.0821 atm L mol^-1K^-1

            dT = change in temperature

Putting the values in equation 1-

            1.75 atm x d(V) = 2.50 mol x (0.0821 atm L mol^-1K^-1) x 125.0 K

            Or, 1.75 atm x d(V) = 25.65625 atm L

            Or, d(V) = 25.65625 atm L / 1.75 atm

            Or, d(V) = 14.661 L

Hence, resultant increase in volume = 14.661 L

# Work done during expansion is given by, W = - p (dV)        - at constant pressure

            Or, W = - 1.75 atm x (14.661 L)

Or, W = - 25.65675 atm L                                       ; [1 atm L = 101.33 J]

            Or, W = - 25.65675 x (101.33 J) = 2599.7984775 J

Hence, work done = - 2599.80 J                          

Note: the –ve sign of (- 2599.80 J) indicates that work is done BY the system (gas) during expansion.