Answer:
For a perfect gas, \[\frac{PV}{T}=\]constant \[(K)\] Given \[\text{P}{{\text{V}}^{\text{2}}}=\]constant \[\left( K' \right)\] \[\therefore \] \[P=\frac{K'}{{{V}^{2}}}\] \[\therefore \] \[\frac{K'}{{{V}^{2}}}.\frac{V}{T}=K\] or \[VT=\frac{K'}{K}=\]constant Hence \[V\propto \frac{1}{T}\] That is the expansion of the gas will result in the decrease of temperature.
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