TS EAMCET · Physics · Kinetic Theory of Gases
If the heat required to increase the rms speed of 4 moles of a diatomic gas from \(v\) to \(\sqrt{3} \mathrm{v}\) is 83.1 kJ, then the initial temperature of the gas is (Universal gas constant \(=8.31 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\) )
- A \(377^{\circ} \mathrm{C}\)
- B \(327^{\circ} \mathrm{C}\)
- C \(227^{\circ} \mathrm{C}\)
- D \(277^{\circ} \mathrm{C}\)
Answer & Solution
Correct Answer
(C) \(227^{\circ} \mathrm{C}\)
Step-by-step Solution
Detailed explanation
\(v_{rms} \propto \sqrt{T} \implies T \propto v_{rms}^2\) \(T_f = (\frac{\sqrt{3}v}{v})^2 T_i = 3T_i\) \(\Delta T = T_f - T_i = 3T_i - T_i = 2T_i\) \(Q = n C_v \Delta T\) \(C_v = \frac{5}{2}R\) (for diatomic gas) \(Q = n (\frac{5}{2}R) (2T_i) = 5nRT_i\) \(T_i = \frac{Q}{5nR}\)…
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