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MHT CET · Chemistry · States of Matter

Which of the following equations gives combined relationship of Boyle's law and Charle's law?

  1. A \(\frac{\mathrm{P}_1 \mathrm{~V}_2}{\mathrm{~T}_1}=\frac{\mathrm{P}_2 \mathrm{~V}_1}{\mathrm{~T}_2}\)
  2. B \(\mathrm{n}=\frac{\mathrm{RT}}{\mathrm{PV}}\)
  3. C \(\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}\)
  4. D \(\mathrm{p}=\frac{\mathrm{RT}}{\mathrm{nV}}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}\)

Step-by-step Solution

Detailed explanation

Charles' law is the change in volume with respect to temperature at constant pressure, while Boyle's law is the change in volume with respect to pressure at a constant temperature.
Taking a gas of volume \(V_1\), pressure \(P_1\) and temperature \(T_1\), and let its change have a state \(\left(\mathrm{V}_2, \mathrm{P}_2, \mathrm{~T}_2\right)\), then according to Boyle's law,
\(P_1 V_1=P_2 V_2 \ldots \ldots . . . .(1)\)
Then keeping this constant pressure, move to state \(\left(\mathrm{V}_2, \mathrm{P}_2, \mathrm{~T}_2\right)\) using Charles' law,
\(\frac{\mathrm{V}_2}{\mathrm{~T}_1}=\frac{\mathrm{V}_2}{\mathrm{~T}_2} \ldots \ldots . . . .(2)\)
Solving for \(V_2\), in both (1) and (2) and then equating them, we get-
\(\frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{~T}_1}=\frac{\mathrm{P}_2 \mathrm{~V}_2}{\mathrm{~T}_2}\)
After changing the pressure, volume and temperature of the gas, still their product is equal, suggesting that the relation is constant: \(\frac{P V}{T}=k\).