A.) A piston has an external pressure of 5.00 atm. How much work has been done i
ID: 953527 • Letter: A
Question
A.) A piston has an external pressure of 5.00 atm. How much work has been done in joules if the cylinder goes from a volume of 0.170 liters to 0.570 liters?
B.) An ideal gaseous reaction (which is a hypothetical gaseous reaction that conforms to the laws governing gas behavior) occurs at a constant pressure of 35.0 atm and releases 73.8 kJ of heat. Before the reaction, the volume of the system was 7.00 L . After the reaction, the volume of the system was 3.00 L .
Calculate the total internal energy change, E, in kilojoules.
C.)
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 4.40 to 2.20 L . When the external pressure is increased to 2.50 atm, the gas further compresses from 2.20 to 1.76 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 4.40 to 1.76 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?
D.) A volume of 120. 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 21.40 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)
E.) The specific heat of water is 4.18 J/(gC). Calculate the molar heat capacity of water.
Explanation / Answer
) A piston has an external pressure of 5.00 atm. How much work has been done in joules if the cylinder goes from a volume of 0.170 liters to 0.570 liters?
Work ; dw = pdV
Here p = 5.00 atm or 506625 Pa
dV = V2-V1= 0.570 L- 0.170 L= 0.400 L
1 liter = 0.001 m^3
SO 0.400 L* 0.001 m^3/ 1.0 L=4*10^-4 m^3
Work = pdV
= 506625 Pa*4*10^-4 m^3
=202.65 Pa m^3
= 202.65 J
B.) An ideal gaseous reaction (which is a hypothetical gaseous reaction that conforms to the laws governing gas behavior) occurs at a constant pressure of 35.0 atm and releases 73.8 kJ of heat. Before the reaction, the volume of the system was 7.00 L . After the reaction, the volume of the system was 3.00 L .
Calculate the total internal energy change, E, in kilojoules.
The change in internal energy that accompanies the transfer of heat, q, or work, w:
dE = q+ w
Here q= 73.8 Kj
W = Work ; dw = pdV
Here p = 35.00 atm or 3.55*10^6Pa
dV = V2-V1= 3.00 L- 7.00 L= -4.00 L
1 liter = 0.001 m^3
SO 4.00 L* 0.001 m^3/ 1.0 L=4*10^-3 m^3
Work = pdV
= 3.55*10^6Pa *- 4*10^-3 m^3
=-14200 Pa m^3
= - 14200 J or -14.2 KJ
dE = q+ w
dE =73.8 Kj -14.2 KJ
dE = 59.6 KJ.
For c (Figure 2) not given
D.) A volume of 120. 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 21.40 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)
Q= mc*dT
For water :
Here dT= 21.40-22.00= 0.60
Mass of water is equal to the volume of water because it s density = 1.0 g/mL
Q for water = 120.0 *4.18 * 0.6
= 300.96 J
This amount of heat is released bt steel rod:
Q steel = mc*dT
Here dt = 21.40-2.00= 19.4
300.96 J= m*0.452 J/(gC) *19.4C
m=34.32 g
) The specific heat of water is 4.18 J/(gC). Calculate the molar heat capacity of water.
To calculate the molar heat capacity of water we will use molar mass of water:
Molar mass of H2O =18.02 g/ mol
Here The specific heat of water is 4.18 J/(gC).
4.18 J/(gC) *18.02 g/mol H2O
= 75.32 J/C
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