According to Charles' law

{(d V / d T)}_{P} = K

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{(d V / d T)}_{P} = - K

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{(d V / d T)}_{P} = - K / T

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None of these

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Solution

The correct option is A ${(d V / d T)}_{P} = K$
According to Charles' law, $V = K T$ . On differentiating at constant pressure we get,
${(\frac{d V}{d T})}_{P} = K$

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Similar questions

(d V / d T)_{p} = - K

(d V / d T)_{p} = K

Boyle's law may be represented as

1.[dp/dv]T=K/V

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

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

4.[dp/dv]T=K/V^2

{(\frac{d E_{c e l l}}{d T})}_{p} > 0

△ S

△ S = n F T {(\frac{d E}{d T})}_{p}

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