A cylinder contains 0.250mol 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 1.00atm 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. Part A How much work is done by the gas in this process? Part B On what is this work done?Part C What is the change in internal energy of the gas?Part D How much heat was supplied to the gas? Part E How much work would have been done if the pressure had been 0.50atm?

We are given a cylinder contains CO2 gas with a number of moles n = 0.250 mol at an initial temperature T_1 = 27 C° which in Kelvins equals 300 K when the pressure is constant and equals p = 1 atm.Then the gas heated up to temperature T_2 = 127C° which in Kelvins equals 400 K.

a) see the attached graph

b) The pressure is constant so the work done at a constant pressure is

W = pΔV (1)

We are not given ΔV ! So from ideal gas law

pV = nRT

As the volume V change, the temperature T will change and the ideal gas law for this change will be

pΔV=nRΔT

ΔV=nR/p( T_2- T_1)

Now we can sub in equation (1) by the value of dV and calculate the work done W where

W = pΔV

= p*nR/p( T_2- T_1)

= 208 J

c ) The work in part (b) is positive which leads us to conclude that the gas applies a work on the piston.

d - The temperature change from 300 K to 400 K and as we know, the internal energy depends on the temperature T. So the change in internal energy ΔU would express by

ΔU=nCvΔT

ΔU=nCv(T_2- T_1)

Where Cv is the molar heat specific capacity at constant volume and for CO2, Cv = 28.46 J/mol K

Now we can sub by the values of Cv , T_2 T_1and n to calculate ΔU

ΔU=nCv(T_2- T_1)

=712 J

e ) The heat Q that must be supplied to the gas to increase its internal energy and the work done is given by the first law of thermodynamics where

Q = ΔU + W

Now sub by the value of ΔU that we calculated in part (d) and the work done that we calculated in part (b) to find the added heat Q

Q = ΔU + W

= 920 J

f) In part (b) we calculated the work done for p = 1 atm which is constant during the process and from the equation W = nR(T_2- T_1) we will conclude that the work does not depend on the pressure in this case, therefore, the work done will not change as the pressure change and it will be the same

W= 208 J

You can repeat the steps in part (b) to confirm the result.