One mole of an ideal monoatomic gas (gamma = 5/3) expands adiabatically from volume V to 8 V. If initial temperature is 300 K, what is the final temperature?
Answer and explanation
Correct answer: C. 75 K
- Step 1: Adiabatic relationship: T1 V1^(gamma - 1) = T2 V2^(gamma - 1). Step 2. gamma - 1 = 5/3 - 1 = 2/3. Step 3. T2 = T1 (V1 / V2)^(2/3) = 300 (1/8)^(2/3). Step 4. (1/8)^(2/3) = ((1/8)^(1/3))^2 = (1/2)^2 = 1/4. Step 5. T2 = 300 (1/4) = 75 K. Answer: 75 K. Why not the others:* 150 K uses (1/8)^(1/3) = 1/2 without squaring; 37.5 K uses 1/8.
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