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A)An ideal gaseous reaction (which is a hypothetical gaseous reaction that confo

ID: 475413 • Letter: A

Question

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 45.0 atm and releases 55.4 kJ of heat. Before the reaction, the volume of the system was 6.60 L . After the reaction, the volume of the system was 2.60 L .

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

B)

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 5.40 to 2.70 L . When the external pressure is increased to 2.50 atm, the gas further compresses from 2.70 to 2.16 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 5.40 to 2.16 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?

Express your answer with the appropriate units.

Explanation / Answer

(A) According to first law of thermodynamics,

E = q + W ---------------- > (1)

It is given that,

amount of heat released, q = - 55.4 kJ

Change in volume, V = 2.60 - 6.60 = - 4.00 L

External pressure, P = 45.0 atm

Work done, W = - P V

W = - 45.0 ( - 4.00 )

W = 180.00 L.atm

But, 1 L atm = 0.101325 kJ

So, W = 180.00 * 0.101325

W = 18.24 kJ

Therefore,

E = - 55.4 + 18.24

E = - 37.16 kJ

(B)

SInce temperature is same for both the processes,

q = - W

q = PV -----------(1)

For two step process,

q = 2.00 (2.70 - 5.40) + 2.50 (2.16 - 2.70)

q = -6.75 L.atm

q = - 0.684 kJ

For one step process,

q = 2.50 ( 2.16 - 5.40)

q = - 8.1 L.atm

q = - 0.821 kJ

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