- #1
anisotropic
- 59
- 0
PROBLEM:
One mole of an ideal monatomic gas initially at 300 K (T1) and a pressure of 15.0 atm (P1) expands to a final pressure of 1.00 atm (P2). The expansion occurs via an adiabatic and reversible path. Calculate q, w, ΔU, and ΔH.
SOLUTION:
q = 0 (adiabatic; no heat exchange occurs)
Thus, ΔU = w = CvΔT
ΔH = CpΔT
ΔT = T2 - T1
Need T2 to calculate values (not an isothermal expansion)...
Known equation for reversible adiabatic process: P1V1γ = P2V2γ (γ = gamma)
Using PV = nRT, substitute in and simplify to obtain expression for T2...
T2 = T1(P1/P2)((1-γ)/γ)
*where problems arise...
γ = ?
The solutions manual says γ = 5/3, but I don't know how this value is obtained.
I do know that, as an ideal monatomic gas, the internal energy (U) of the gas (per mole?) is: 3/2RT (we assume only translations occur). But how is this related to Cv, Cp, and more importantly, γ?
One mole of an ideal monatomic gas initially at 300 K (T1) and a pressure of 15.0 atm (P1) expands to a final pressure of 1.00 atm (P2). The expansion occurs via an adiabatic and reversible path. Calculate q, w, ΔU, and ΔH.
SOLUTION:
q = 0 (adiabatic; no heat exchange occurs)
Thus, ΔU = w = CvΔT
ΔH = CpΔT
ΔT = T2 - T1
Need T2 to calculate values (not an isothermal expansion)...
Known equation for reversible adiabatic process: P1V1γ = P2V2γ (γ = gamma)
Using PV = nRT, substitute in and simplify to obtain expression for T2...
T2 = T1(P1/P2)((1-γ)/γ)
*where problems arise...
γ = ?
The solutions manual says γ = 5/3, but I don't know how this value is obtained.
I do know that, as an ideal monatomic gas, the internal energy (U) of the gas (per mole?) is: 3/2RT (we assume only translations occur). But how is this related to Cv, Cp, and more importantly, γ?
Last edited: