Ideal Gas Expansion: Reversible Process Calc.

[nancy awwad]

One mole of an ideal gas having CV = 25.0 J.K−1 is in a volume V1 = 28.0 L at P1 =
1.00 bar. It is expanded reversibly along a path given by P = P1 − 0.005000(V − V1) to
a final volume V2 = 100.0 L. Calculate the final temperature T2 and the values of q,
w, ΔE, ΔH for this change in state.

 

[I like Serena]

Welcome to PF, nancy_awwad!

It would help if you would show some of what you tried.
Then we could better point you in the right way.

Let's start with P2.
Can you calculate it from the formula you already have?

 

[nancy awwad]

we already know that PV=nRT then i ve calculated T1

 

[I like Serena]

we already know that PV=nRT then i ve calculated T1

Good!
So what is T1?

To find T2 you first need to find P2.

Can you calculate P2 using P = P1 − 0.005000(V − V1)?

 

[DeeNos]

I would like to start by clarifying that the ideal gas expansion described in this question is a reversible process, meaning that the system can be returned to its initial state without any change in its surroundings. In this scenario, the expansion of one mole of an ideal gas with a specific heat capacity of 25.0 J.K−1 is being studied. The gas is initially at a volume of 28.0 L and a pressure of 1.00 bar, and it is expanded along a reversible path characterized by the equation P = P1 − 0.005000(V − V1). The final volume of the gas is 100.0 L.

To calculate the final temperature T2, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. We can rearrange this equation to solve for T, giving us T = PV/nR. Substituting the values given in the question, we get T2 = (1.00 bar)(100.0 L)/(1 mol)(8.314 J.K−1.mol−1) = 120.3 K.

To determine the values of q, w, ΔE, and ΔH for this change in state, we can use the first law of thermodynamics, which states that ΔE = q + w, where ΔE is the change in internal energy of the system, q is the heat transferred, and w is the work done on or by the system.

To calculate q, we need to use the equation q = nCΔT, where C is the molar heat capacity and ΔT is the change in temperature. In this scenario, the change in temperature is T2 - T1 = 120.3 K - 298 K = -177.7 K. Substituting the values given in the question, we get q = (1 mol)(25.0 J.K−1)(-177.7 K) = -4442.5 J.

To determine the work w, we can use the equation w = -∫PdV, where P is the pressure and V is the volume. Substituting the given equation for P and integrating it from V1 to V2, we get w = -∫(P1