Question

Pressure versus temperature graph of an ideal gas is shown, The density of gas at point $A$ is ${\rho }_0$ then the density of gas at point $B$ would be

• A. $\frac{3}{4}{\rho }_0$
• B. $\frac{3}{2}{\rho }_0$
• C. $\frac{4}{3}{\rho }_0$
• D. $2{\rho }_0$

Answer 1

Answer: (B)

If mass of a gas is constant then there is variation in density of gas with the variation of temperature and pressure because the molar volume of the gas goes on changing because $\rho =\frac{M}{V}$
We know, from Ideal gas law $p=\frac{\rho RT}{M}$
For point $A$ , $p=\frac{{\rho }_{0}R{T}_0}{M}$ $\Rightarrow M=\frac{{\rho }_{0}R{T}_0}{p}$ ...(i)
For point $B$ , $3p=\frac{\rho R(2T_0)}{M}=\frac{\rho R(2T_0)}{{\left(\rho \right)}_0{\left(RT\right)}_0}\times p$ [By Eq. (i)]
$\Rightarrow \rho =\frac{3{\rho }_0}{2}$