Question

Isothermally and reversibly one mole of neon expands from 2 m³ to 20 m³ and produces 831.4 J of work. The temperature at which expansion takes place is ____________.
(R = 8.314 J K^-1 mol^-1)

पर्याय

• 434.2 K

• 4342 K

• 43.42 K

• 316.42 K

Answer 1

Isothermally and reversibly one mole of neon expands from 2 m³ to 20 m³ and produces 831.4 J of work. The temperature at which expansion takes place is 43.42 K.

Explanation:

W_max = −2.303 nRT log₁₀ `"V"_2/"V"_1`

−831.4 J = −2.303 × 1 mol × 8.314 J K⁻¹ mol⁻¹ × T × log `(20/2)`

T = `(-831.4)/(-2.303 xx 8.314)` K = 43.42 K