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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V \propto T

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Solution

The correct options are
A $(d V / d T)_{p} = K$
D $V \propto T$
Charles' law : The volume of a given gas is proportional to the absolute temperature.
$V \propto T ∴ {[\frac{\partial V}{\partial T}]}_{P} = K$

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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

d ϕ = f (T) d T + (T / V) d V a n d d ψ = T d p + (T / p^{2}) d V,

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