Question

consider the reversible isothermal constant preasure freezing of 1.00 mol of water at 0.00C and 1.00 atm, calculate q,w, delt u, delt H

Answer 1

∆U = 0, ∆H = 0 , q = w, w = 2269.722 ln (V2/V1).

Step-by-step explanation:

This question is about the application of the first law of thermodynamics. An ideal gas is one that can be represented by the equation of state, PV = nRT. The ideal gas' internal energy is a function of of temperature.

Isothermal changes are changes that occur at constant temperature. At constant temperature, ∆U = 0. Thus we have that q = w = ∫ P dv for isothermal expansion or compression.

Recall that; PV = nRT , thus P = nRT/V. Slotting in the P value into the equation q = w = ∫P dv. We will have w = nRT ln P1/P2.

Thus, q= w=( 1 × 8.314 × 273) ln ( P1/P2).

We are told that the pressure is contsant, yet no volume is given.

Assuming the we have the volume, the formula will be;

q = w = ( 1 × 8.314 × 273) ln (V2/V1).

Because Boyle's law states that P1 V1 = P2V2. Thus, V2 /V1 = P1/P2.