According to Charles' law.

V \propto \frac{1}{T}

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{(\frac{d V}{d T})}_{P} = K

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{(\frac{d T}{d V})}_{P} = K

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{(\frac{1}{T} - \frac{V}{T^{2}})}_{P} = 0

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Solution

The correct options are
B ${(\frac{d V}{d T})}_{P} = K$
C ${(\frac{d T}{d V})}_{P} = K$
According to Charles' law, at constant pressure
$V \propto T$
$V = K T$ or $T = K V$
$\frac{d V}{d T} = K$
$\frac{d T}{d V} = K$

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3.[dp/dv]T=-K/V

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