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, ΔE, 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/(g⋅∘C)

specific heat of steel = 0.452 J/(g⋅∘C)

E.) The specific heat of water is 4.18 J/(g⋅∘C). Calculate the molar heat capacity of water.

Answer 1

) 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, ΔE, 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/(g⋅∘C)

specific heat of steel = 0.452 J/(g⋅∘C)

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/(g⋅∘C) *19.4⋅∘C

m=34.32 g

) The specific heat of water is 4.18 J/(g⋅∘C). 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/(g⋅∘C).

4.18 J/(g⋅∘C) *18.02 g/mol H2O

= 75.32 J/∘C