Question

According to Charles' law (1) $ {{\left( \frac{dV}{dT} \right)}_{p}}=K $ (2) $ {{\left( \frac{dT}{dV} \right)}_{p}}=K $ (3) $ {{\left( \frac{1}{T}-\frac{V}{{{T}^{2}}} \right)}_{p}}=0 $ (4) $ V\propto \frac{1}{T} $

• A. 1, 2 and 3 are correct
• B. 1 and 2 are correct
• C. 2 and 4 are correct
• D. 1 and 3 are correct

Answer 1

Answer: (A)

According to Charles law $ V\propto t $ at constant p $ V=kT $ at constant p Or $ {{\left( \frac{dV}{dT} \right)}_{p}}=K $
Further, $ T\propto V $ $ T=KV $
Or $ {{\left( \frac{dT}{dV} \right)}_{p}}=K $
Further, $ \frac{V}{T}=K $
$ \frac{1}{T}.V=K $
Differentiating w.r.t. $ V $
$ {{\left( \frac{1}{T}-\frac{V}{{{T}^{2}}} \right)}_{p}}=0 $