Question

An ideal gas (which is is a hypothetical gas that conforms to the laws governing gas behavior) confined to a container with a massless piston at the top. (Figure 2)  A massless wire is attached to the piston. When an external pressure of 2.00 atm is applied to the wire, the gas compresses from 5.90 to 2.95 L . When the external pressure is increased to 2.50 atm, the gas further compresses from 2.95 to 2.36 L .

In a separate experiment with the same initial conditions, a pressure of 2.50 atm was applied to the ideal gas, decreasing its volume from 5.90 to 2.36 L in one step.

If the final temperature was the same for both processes, what is the difference between q for the two-step process and q for the one-step process in joules?

part A:

An ideal gaseous reaction (which is a hypothetical gaseous reaction that conforms to the laws governing gas behavior) occurs at a constant pressure of 50.0 atm and releases 63.1 kJ of heat. Before the reaction, the volume of the system was 9.00 L . After the reaction, the volume of the system was 2.60 L .

Calculate the total internal energy change, ΔE, in kilojoules.

Part B:

Answer 1

delta U = q+w
q1+w1 = q2+w2
This allows us to use sample algebra rules to state that
w1-w2 = q2-q1
W1 = -pdv
=- 2.5(2.36-5.9)
= 8.85 atm-L
1 atm = 101 joules
= 8.85*101 =893.85 joules
Then repeat the process for the two step process
W2 = step1 + step2
   = -Pdv+ -Pdv
   = -2*(2.95-5.9) -2.5(2.36-2.95)
   = 5.9 +1.475
   = 7.375 atm- lit
   = 7.375*101 = 744.875 joules
w1-w2 = q1-q2
     = 893.85-744.875
     = 148.975 joules
2 . dU = dQ-W
       = dQ-Pdv
       = 63100-50(9-2.6)*101 joule
       = 63100-32320
       = 3078 joules >>>> answer
inter nal energy is 3078 joules
                     3.078 kj >>>> answer