Question

An ideal gas is initially at temperature T and Volume V .If volume increases by ∆V due to increase in temperature of ∆T,pressure remaining constant . The quantity p = [tex] \frac{dV}{TdT} varies with temperature. What is the nature of graph of p Vs T(Temperature).

Where, dV = ∆V​

Answer 1

Correct question :- An ideal gas is initially at temperature T and Volume V .If volume increases by ∆V due to increase in temperature of ∆T,pressure remaining constant . The quantity p = $\frac{dV}{TdT}$ varies with temperature. What is the nature of graph of p Vs T(Temperature).

Where, dV = ∆V

Solution

Given

An ideal gas initially at temperature T and Volume V. Thereafter, these two quantities are increases keeping pressure constant.

To Find

The nature of plot of The quantity p = $\frac{dV}{TdT}$ Versus temperature

Concepts

In simple terms, we will solve it without using differentiation.

Charle's law

This law states that volume of gas in a container is directly proportional to Temperature, pressure being constant.

Hence, $\frac{V}{T}$ = Constant (Say K)

Solution

We have, $\frac{V}{T}$ = K

=> $\frac{dV}{dT}$ = K

Now,

=> $\frac{dV}{TdT}$ = $\frac{1}{KT}$

=> P = $\frac{1}{KT}$

=> P T = K.

This, equation resembles xy = K, Thus the plot is rectangular hyperbola.