1. An ideal gas (which is is a hypothetical gas that conforms to the laws govern
ID: 891835 • Letter: 1
Question
1.
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 6.60 to 3.30 L . When the external pressure is increased to 2.50 atm, the gas further compresses from 3.30 to 2.64 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 6.60 to 2.64 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?
2. A volume of 110. mL of H2O is initially at room temperature (22.00 C). A chilled steel rod at 2.00 C is placed in the water. If the final temperature of the system is 21.50 C , what is the mass of the steel bar?
Use the following values:
specific heat of water = 4.18 J/(gC)
specific heat of steel = 0.452 J/(gC)
3. The specific heat of water is 4.18 J/(gC). Calculate the molar heat capacity of water.
Explanation / Answer
1. As you are asking about changes in volumes I assume q as work done within the system.
W=PDeltaV
Two steps process
a. W=202650Pa*(6.6E-03 - 3.3E-03)L= 668.745 J
2. Qwater=QSteel
mwaterCpwater(T2-T1)=msteelCpsteel(T2-T1)
Solving for msteel, msteel=26.08g Steel bar.
3. 4.18 J/(g*°C) * (18g /mol) = 75.24 J/(mol*°C)
b. W=50662.5Pa*(3.3E-03 - 2.64E-03)L= 33.43725J
total = 702.18225J
One step process
W= 253312.5Pa*(6.6E-03 - 2.64E-03)L= 1003.1175 J
2.
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