One mole of an ideal diatomic gas expands from volume V to 2V isothermally at a temperature 27°C and does W joule of work. If the gas undergoes same magnitude of expansion adiabatically from 27°C doing the same amount of work W, then its final temperature will be (close to) __________ °C.
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Question

One mole of an ideal diatomic gas expands from volume V to 2V isothermally at a temperature 27°C and does W joule of work. If the gas undergoes same magnitude of expansion adiabatically from 27°C doing the same amount of work W, then its final temperature will be (close to) __________ °C.

Options

G1 Physics · Accepted Answer

For the isothermal expansion, W = nRT ln(V₂/V₁). Here, n = 1, T = 300 K, V₂/V₁ = 2. Therefore, W = 300R ln2 ≈ 300R(0.693). For the adiabatic process, W = nR(T₁-T₂)/(γ-1). For a diatomic ideal gas, γ = 1.4. The problem states that the work done is the same in both processes. Hence, 300R(0.693) = R(300-T₂)/(1.4-1). Cancel R: 300(0.693) = (300-T₂)/0.4. Therefore, 300-T₂ = 83.16. Thus, T₂ = 216.84 K. Convert to Celsius: T₂ = 216.84-273 ≈ -56.2°C. Correct option: B.