Question

1 mole of an ideal gas in a cylindrical container have the P-V diagram as shown in figure. If $V_2$ =$4V_1$, then the ratio of temperatures $\frac{T_1}{T_2}$ will be:

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{3}{2}$
• D. $\frac{3}{4}$

Answer 1

The correct option is A $\frac{1}{2}$

Ideal gas equation is given by,

$PV=nRT$

or

$T=\frac{PV}{nRT}$ ...........(1)

Given,

$P{V}^{1/2}=A$

Multiplying both side by $\sqrt{V}$

$PV=A\sqrt{V}$ .........(2)

Substituting eqn (2) in eqn (1), we get,

$T=\frac{\sqrt{V}}{nR}$

$⟹T\propto \sqrt{V}$

Now,

$\frac{T_1}{T_2}=\sqrt{\frac{V_1}{V_2}}=\sqrt{\frac{V_1}{4V_1}}=\frac{1}{2}$

So,

$\frac{T_1}{T_2}=\frac{1}{2}$