An ideal gaseous reaction occurs at a constant pressure of 40.0 \\({\ m atm}\\)

Question

An ideal gaseous reaction occurs at a constant pressure of 40.0({ m atm}) and releases 56.7({ m kJ}) of heat. Before the reaction, the volume of the system was 8.80({ m L}) . After the reaction, the volume of the system was 2.40({ m L}) .

Calculate the total internal energy change, ( exttip{Delta U}{Delta U}), in kilojoules.

An ideal 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 ( m atm) is applied to the wire, the gas compresses from 4.90 to 2.45({ m L}) . When the external pressure is increased to 2.50 ( m atm), the gas further compresses from 2.45 to 1.96({ m L}) .

In a separate experiment with the same initial conditions, a pressure of 2.50 ( m atm) was applied to the ideal gas, decreasing its volume from 4.90 to 1.96({ m L}) in one step.

If the final temperature was the same for both processes, what is the difference between ( exttip{q}{q}) for the two-step process and ( exttip{q}{q}) for the one-step process in joules?

Explanation / Answer

deltaH=delta U +PdeltaV

where H is enthalpy

U is internal energy

P is pressure and v is volume

now using given data in the eq. written above

-56.7*1000=delta U +40(8.80-2.40)

-56700-256=delta U

delta U=-56.95KJ

second part is not clear but i think u can solve by applying the same fomula

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