Question:
In the following P-V diagram of an ideal gas, two adiabates and two isotherms at T₁=300K and T₂=200K are shown. Values are VA=2, VB=8, VC=16. Find VD.
In the following P-V diagram of an ideal gas, two adiabates and two isotherms at T₁=300K and T₂=200K are shown. Values are VA=2, VB=8, VC=16. Find VD.
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Adiabatic curves are steeper than isotherms.
Updated On: Mar 20, 2026
- 4 unit
- <4 unit
- >5 unit
- 5 unit
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The Correct Option is B
Solution and Explanation
Step 1: For adiabatic process: TV^γ-1 = constant
Step 2: 300 × 2^γ-1 = 200 × VD^γ-1
Step 3: Solving gives VD < 4.
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