Answer:
a) see the attached graph
b) 208 J
c) the gas applies a work on the piston.
d) 712 J
e) 920 J
f) 208 J
Explanation:
Given
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.
