Q8 / 160 · TS EAMCET 2023 (12 May Shift 1) · Chemistry

If 2.5 moles of an ideal gas at a certain temperature are allowed to expand isothermally and reversibly from an initial volume of $2 \mathrm{dm}^3$ to $20 \mathrm{dm}^3$, the work done by the gas is $-16.5 \mathrm{~kJ}$. The temperature (in $\mathrm{K}$ ) of the gas is (Round off to the nearest value) $\left(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)$

  • A
    $445$
  • B
    $245$
  • Correct answer
    C
    $345$
  • D
    $745$

Solution

Reversible isothermal work is given by :$\mathrm{W}=-2.303 \mathrm{nRT} \log \left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)$ $\begin{aligned} & =-2.303(2.5)(8.314)(\mathrm{T}) \log \left(\frac{20}{2}\right) \\ & =-47.87 \mathrm{~T} \\ & =-16500 \mathrm{~J} \\ & \Rightarrow \mathrm{T}=344.68 \cong 345 \mathrm{~K}\end{aligned}$

Hence, the correct answer is option: 'C'.

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