TS EAMCET · Chemistry · Thermodynamics (C)
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, \(\left.\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)(1 \mathrm{~L} \mathrm{~atm}=101.3 \mathrm{~J})\)
- A \(7.476\)
- B \(11.214\)
- C \(3.738\)
- D \(14.952\)
Answer & Solution
Correct Answer
(A) \(7.476\)
Step-by-step Solution
Detailed explanation
\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}…
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