Thermal Physics - Isothermal vs Adiabatic

An ideal monatomic gas expands quasi-statically to twice its volume. If the process is isothermal, the work done by the gas is W_i. If the process is adiabatic, the work done by the gas is W_a. Which is the following is true?

A. W_i = W_a
B. 0 = W_i  W_a 
C. 0  W_i  W_a 
D. 0 = W_a  W_i
E. 0  W_a  W_i

(GR0177 #06)

Solution:

Isothermal and Adiabatic P-V Diagram

• Adiabatic connects high-T isotherm and low-T isotherm.
• Isothermal line is always higher than the adiabatic line and they both end at the same volume
• The area under the isothermal line is bigger than the adiabatic → W_a  W_i

Answer: E

Calculation:

Isothermal:

PV = constant
P_i V_i  = P_f V_f  = constant
Given V_f  = 2V_i

P_i V_i  = 2P_f V_i

P_f (iso) = ½ P_i

Adiabatic:

PV^γ = c
P_i V_i^γ = P_f V_f ^γ = constant
V_f  = 2V_i
P_iV_i^γ  = P_f (2V_i)^γ = 2^γP_f V_i^γ 
P_f (adi) = (1/2^γ)P_i 
  
Therefore,

P_{f (iso)} /P_{f (adi)} = 2^γ / 2
P_f (iso) = 2^γ−1P_{f (adi)}

→ P_adi  P_iso

W = ∫ PdV → W_adi  W_iso