WBJEE · Chemistry · States of Matter
For same mass of two different ideal gases of molecular weights \(M_{1}\) and \(M_{2}\), Plots of log \(V\) vs log \(p\) at a given constant temperature are shown. Identify the correct option.
- A \(M_{1}>M_{2}\)
- B \(M_{1}=M_{2}\)
- C \(M_{1} < M_{2}\)
- D Can be predicted only if temperature is known
Answer & Solution
Correct Answer
(A) \(M_{1}>M_{2}\)
Step-by-step Solution
Detailed explanation
For ideal gas, \(\log \frac{k}{M_{2}}, \log \frac{k}{M_{1}}\) \(\because\) \(p V=\frac{W}{M} R T\) Let, \(W R T=K\) Taking log of both the sides of Eq. (i) \[ \log p+\log V=\log \frac{K}{M} \] or \(\log V=-\log p+\log \frac{K}{M}\) [On compairing Eq. (ii) with \(y=m x+C]\)…
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