[thetada]

For the following question:

Of an ideal gas put in an isolated cylinder, the initial volume is 4 liters; the pressure is 3 atmospheres; and the temperature is 20⁰C. This gas has been exposed to adiabatic transformations until the volume decreases to 2 liters. Calculate the new pressure, temperature, the work done and change in internal energy. (The question doesn't specify so I've assumed that it's a monoatomic gas, hence with γ = 5/3)

PV^γ = constant

So P₁V₁^γ = P₂V₂^γ

P₂ = 303975 x 0.004^(5/3) / 0.002^(5/3) = 965060.470Pa (3dp)

Using the ideal gas law n = P₁V₁/RT₁ = 303975 x 0.004 / 8.314 x 293.15 = 0.4989mol

P₁V₁/T₁ = P₂V₂/T₂

So T₂ = P₂V₂T₁/P₁V₁

= 465.347K

Then, using the following equation:

W = -ΔU = -3/2nRΔT = -1.5 x 0.4989 x 8.314 x (465.347-293.15)

W = -1071.370J

ΔU = 1071.370J

I know I've been a little sloppy by basing subsequent calculations on prematurely rounded intermediary figures, but I just need to check I'm not way off base. To provide some context, I'm ensuring I can solve problems printed in an eduational study of preservice physics teachers, since I am responsible for producing materials for preservice chemistry teachers. I'd really appreciate it if anyone could point out any errors I've made.

Thank you

[Enthalpy]

Hi thetada!

For most questions, like the new pressure, you do need γ. Either you give algebraic answers, or you must suppose some γ as you did, since the questions give numerical values. Monoatomic, or diatomic like air, or anything else.

PV^γ=const is convenient between P and V. Other formulas are more direct for other variables, they hold for adiabatic transformations of perfect gases and result from the combination with PV/T=const:

P=T^C_p/R×const
V=T^C_v/R×const
where C_v/R is 3/2 for a monoatomic gas (3 translations) and 5/2 for a diatomic one without vibrations (adds 2 rotations), and C_p/R = 1 + C_v/R, which is the same as H=U+PV.
To remember it: C_p gives P and C_v gives V.

The change of internal energy is ΔU=ΔT×C_v
and of enthalpy ΔH=ΔT×C_p which gives you the work.

Most often, the temperature ratio is best computed first, and everything else deduced from it, first as ratios.

[thetada]

Hi Enthalpy,

Thanks for that, appreciate it.

Just to clarify, is it that my numbers are fine but there's a quicker way to calculate them? I'll play around with the other equations you've shared anyway, make sure all roads lead to Rome =)

Thanks!