Question

A gas is confined to a container with a massless piston at the top. A massless wire is attached to the piston. When an external pressure of 2.03 bar is applied to the wire, the gas compresses from 6.40 to 3.20 L . When the external pressure is increased to 2.53 bar, the gas further compresses from 3.20 to 2.56 L .

In a separate experiment with the same initial conditions, a pressure of 2.53 bar was applied to the gas, decreasing its volume from 6.40 to 2.56 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?

Answer 1

q in case of two -step process will be less in magnitude than q in case of one step process.

Explanation:

Heat q given to a system is used in 1) increasing the internal energy of the gas

and 2) work done in increasing the volume . If there is decrease in volume , work done is negative.

In both the cases change in internal energy of the gas is same as the rise in temperature is same. But work done which is negative in nature will have greater value in one step process. It is so because work done is equal to external pressure multiplied by change in volume. Change in volume is same but external pressure is more in case of two step process. Hence negative work done is more in case of two step process. Hence q in case of two step process will be lesser.

q = change in internal energy - work done on the gas.

Answer 2

Note that 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:
∆U₁ = ∆U₂
The change in internal energy of the gas equals the heat added to the gas plus work done on it:
Hence,
Q₁ + W₁ = Q₂ + W₂
So the difference in heat transfer between the two process is:
∆Q = Q₂ - Q₁ = W₁ - W₂

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)


I hope my guide has come to your help. Have a nice day ahead and may God bless you always!