JEE Advanced 2021 Paper 2 · Q08 · Kinetic Theory of Gases

Correct answer:10.00

Solution

Isothermal compression of the trapped air (ideal gas): $p_0 v_0 = (p_0 + \Delta p)(v_0 - \Delta v)$, i.e. $10^5 \times 3.3 = (10^5 + \Delta p)(3.3 - 0.3) = (10^5 + \Delta p)(3.0)$. So $10^5 + \Delta p = 3.3\times 10^5/3 = 1.1\times 10^5$ Pa, giving $\Delta p = 0.1\times 10^5 = 10\times 10^3$ Pa. With $\Delta p = Y\times 10^3$ Pa, $Y = 10.00$.

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