JEE Mains · Physics · STD 11 - 12 . kinetic theory of gases
Initially a gas of diatomic molecules is contained in a cylinder of volume \(V _{1}\) at a pressure \(P_{1}\) and temperature \(250\, K\). Assuming that \(25 \%\) of the molecules get dissociated causing a change in number of moles. The pressure of the resulting gas at temperature \(2000\, K ,\) when contained in a volume \(2 V _{1}\) is given by \(P _{2}\). The ratio \(\frac{P _{2}}{ P _{1}}\) is.
- A \(5\)
- B \(10\)
- C \(13\)
- D \(9\)
Answer & Solution
Correct Answer
(A) \(5\)
Step-by-step Solution
Detailed explanation
\(P V = n R T\) \(P _{1} V _{1}= nR\times 250\) \(P _{2}\left(2 V _{1}\right)=\frac{5 n }{4} R \times 2000\) Divide \(\frac{ P _{1}}{2 P _{2}}=\frac{4 \times 250}{5 \times 2000}\) \(\frac{P_{1}}{P_{2}}=\frac{1}{5}\) \(\frac{P_{2}}{P_{1}}=5\)
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