Q9 / 160 · TS EAMCET 2024 (11 May Shift 1) · Chemistry

At $300 \mathrm{~K}, 3.0$ moles of an ideal gas at 3.0 atm pressure is compressed isothermally to one half of its volume by an external pressure of 6.0 atm . The work done (in kJ ) is $\left(\right.$ Given, $\left.\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)(1 \mathrm{~L} \mathrm{~atm}=101.3 \mathrm{~J})$

  • Correct answer
    A
    $7.476$
  • B
    $11.214$
  • C
    $3.738$
  • D
    $14.952$

Solution

$\begin{aligned} & \text {For ideal gas, } \\ & \text {Mole }=3 \\ & \text {Temp }=300 \mathrm{~K} \\ & \text {Pressure }=3 \mathrm{~atm} \\ & \text {PV }=\mathrm{nRT} \\ & \mathrm{V}=\frac{\mathrm{nRT}}{\mathrm{P}}=\frac{3 \mathrm{~mole} \times 0.082 \mathrm{c} \mathrm{atm} \mathrm{K}^{-1} \mathrm{~mol}^{-1} \times 300 \mathrm{~K}}{3 \mathrm{~atm}} \\ & \mathrm{~V}_1=24.6 \mathrm{~L} \\ & \therefore \mathrm{~V}_2=\frac{\mathrm{V}_1}{2}=\frac{24.6}{2}=12.3 \\ & \Delta \mathrm{~V}=\mathrm{V}_2-\mathrm{V}_1 \\ & =12.3-24.6 \\ & \Delta \mathrm{~V}=-12.3 \mathrm{~L} \\ & \mathrm{P}_{\mathrm{ext}}=6 \mathrm{~atm} \\ & \mathrm{~W}=-\mathrm{P}_{\text {ext }}(\Delta \mathrm{V}) \\ & =-6 \times(-12.3) \mathrm{atm} \mathrm{L} \\ & =73.8 \mathrm{~atm} \mathrm{~L} \\ & \mathrm{~Win} \mathrm{kJ}=\frac{73.8 \times 101.3 \mathrm{~kJ}}{1000} \\ & \mathrm{~W}=7.476 \mathrm{~kJ} .\end{aligned}$

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

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