Recent content by grangr

I get it. For ideal gas, ΔH = nCvΔT + Δ(nRT) = nCvΔT + nRΔT = nCpΔT, thus Cv + R = Cp. Therefore, Cv = Cp - R = 35.4 - 8.314 [J/mol⋅K] = 27.086. To be honest though, I fail to comprehend the concept of internal degrees of freedom still. The question asks about an ideal gas, supposedly it is not...

In my attempt, I thought Cv = (3/2)R and Cp = (5/2)R, as reasoned below. The kinetic energy (KE) for a gas is 3/2⋅(nRT) (Relevant equation #2). For an ideal gas, now that ΔE = nCvT, and KE = 3/2⋅(nRT) Cv = 3/2⋅R (Relevant equation #3) Therefore, given that ΔH = n(Cv+R)ΔT = n([3/2⋅R]+R)ΔT =...

No, I believe there is not. The problem statement was all that was given.

Thank you for your reply. I see. So in (b), it could be like an isothermal process, where ΔT = 0 (thus ΔE = 0) but Q ≠ 0.

Homework Statement If Cp for an ideal gas is 35.4 J/mol⋅K, which of the following is Cv for this gas? a. 12.5 J/mol⋅K b. 20.8 J/mol⋅K c. 29.1 J/mol⋅K d. 27.1 J/mol⋅K e. 43.4 J/mol⋅K Homework Equations ΔH = ΔE + Δ(PV) = Q + W + Δ(PV), and for ideal gas, ΔH = nCvΔT + Δ(nRT) = nCvΔT + nRΔT =...

Homework Statement Which of the following statement(s) is (are) correct when an ideal gas goes from an initial to a final state in a single process? a. No work is done on or by the gas when the volume remains constant. b. No energy is transferred into or out of the gas as heat transfer when the...

Thank you all, Nguyen Son, rude man, and Perok for your explanation. It is much clearer now. :partytime: (And it's good to know that both b) and d) are correct! :wink:) Indeed, I totally did not recall that cos(x) = cos(-x), or that cos(θ)=sin[θ+(π/2)]. :doh: As it turns out, it wasn't a...

Homework Statement The equation y = A sin(kx - wt + pi/2) is the same as a. y = -A sin(kx - wt + pi/2) b. y = A cos(kx - wt) c. y = -A cos(kx - wt) d. y = -A sin(kx - wt - pi/2) e. y = A sin(kx - wt + (3pi)/2) Homework Equations y = A sin[(2pi)/lamda * x - (2pi)/period * t + (phase constant)]...