A cylinder contains 0.250 mol of carbon dioxide (CO2) gas at a temperature of 27.0∘C. The cylinder is provided with a frictionless piston, which maintains a constant pressure of 1.00 atm on the gas. The gas is heated until its temperature increases to 127.0∘C. Assume that the CO2 may be treated as an ideal gas. (a) Draw a pV-diagram for this process. (b) How much work is done by the gas in this process? (c) On what is this work done? (d) What is the change in internal energy of the gas? (e) How much heat was supplied to the gas? (f) How much work would have been done if the pressure had been 0.50 atm?

The answers to this question and its part is given in details in the explanatory section below. But highlights to the answers are as follows -

[a] please see attachment

[b] -  W = 207.85J

[c] - The work done is primarily on the friction less piston of the gas cylinder.

[d] - 711.5J

[e] - 919.35J

[f] - unchanged

Explanation:

Before kicking off in answering the questions, one needs to extract the relevant information needed. From the question, one can notice that there was an increase in temperature from 27°C to 127°C which in turn happened at a constant pressure of 1.00 atm. Therefore, in accordance to the ideal gas law, the volume of a gas will increase with a corresponding increase in temperature.  

[a]

With the already established point above, the pressure - volume (pV) diagram for the process is as follows;

See attachment to the diagram.

[b]

At constant pressure, the work done by the gas is expressed as -

              ΔV  ------------------------ (i)

where  W = work done

             p = pressure

          ΔV = change in volume

Since ΔV was not given as a parameter in the question, one needs to calculate for this value first. From an ideal gas law, we have -

               pV = nRT -----------------------(ii)

where,

               R = ideal gas constant = 8.314 J/mol.K

               n = 0.250 mol

               T = Temperature

              pΔV = nRΔT

              ΔV = (nR/p) ΔT

Now substituting the value of ΔV in equation (i) we have

             W = p (nR/p) ΔT  ---------------------- (iii)

by mere looking at equation (iii) once will notice that p cancels out each other, the new equation becomes

             W =  nR ΔT

             where  ΔT = ( T2 - T1)

                               T2 = 127°C = 400°K

                               T1 = 27°C = 300°K

                        W = 0.250 mol * 8.314 J/mol.K (400 - 300)

                        W = 207.85J

[c]

The work done is primarily on the friction less piston of the gas cylinder.

[d] Since the internal energy witnessed by the gas depends on the applied temperature, hence, it can be calculated as follows -

                 ΔU = n ΔT -------------------- (iv)

where,

            = molar heat specific capacity at constant volume for Carbon di        oxide = 28.46 J/mol.K

calculating for ΔU we have

                ΔU = 0.250 mol * 28.46 J/mol.K (400 -300)

                      =  711.5J

[e]

To know how much heat was supplied to the gas, we need to apply the first law of thermodynamics -

               Q = ΔU + W --------------------- (v)

                   = 711.5 + 207.85

                   = 919.35J

[f]

Since the work done does not depend on the pressure, therefore, a reduction of the pressure from its constant state of 1.00 atm to 0.50 atm will leave the result gotten for the work done in [b] unchanged.