A) A gaseous reaction occurs at a constant pressure of 50. atm and releases 74.2

Question

A) A gaseous reaction occurs at a constant pressure of 50. atm and releases 74.2kj of heat. Before the reaction, the volume of the system was 8.00L After the reaction, the volume of the system was 2.80L

Calculate the total internal energy change in kilojoules.

B) gas is 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 3.30L When the external pressure is increased to 2.50atm the gas further compresses from 3.30 to 2.64atm.

In a separate experiment with the same initial conditions, a pressure of 2.50atm was applied to the gas, decreasing its volume from 6.60 to 2.64L 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 qq for the one-step process in joules?

Explanation / Answer

A).

Delta U = q - P(Delta)V

or Delta U = q + w
(Delta)V = 2.80 - 8.00 = -5.2
since (Delta)V is negative w is positive.

P(delta)V = 5.2 * 50 = 260.
convert to kJ so 260 becomes 26.0
since it release it, its loss of heat so its -74.2 + 26.0 = -48.2kJ

B).

Internal energy is a state function. That means internal energy of the gas can be expressed as function of two state variables, e.g. U = f(T;V). For an ideal gas internal function can be expressed of temperature alone. But is not necessary to make ideal gas assumption to solve this problem.
Because internal energy is a state function, a process changing from state 1 to state 2 has always the same change change in internal energy irrespective of the process design.
The one-step compression and the two two-step compression start at the sam state and end up in same state. the gas undergoes the same change in internal energy:
?U? = ?U?
The change in internal energy of the gas equals the heat added to the gas plus work done on it:
Hence,
Q? + W? = Q? + W?
So the difference in heat transfer between the two process is:
?Q = Q? - Q? = W? - W?

The work done on the gas is given by piston is given by the integral
W = - ? P_ex dV from V_initial to V_final
For constant external pressure like in this problem this simplifies to
W = - P_ex ? ? dV from V_initial to V_final
= P_ex ? (V_initial - V_final)

The work done in one step process is:
W? = 2.5*101325 Pa * (6.60

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