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 $p=\frac{ρRT}{M}$
For point $A$ , $p=\frac{\rho _{0} RT_{0}}{M}$ $\Rightarrow M=\frac{\rho _{0} RT_{0}}{p}$ ...(i)
For point $B$ , $3p=\frac{ρR \left(\right. 2 T_{0} \left.\right)}{M}=\frac{ρR \left(\right. 2 T_{0} \left.\right)}{\left(\rho \right)_{0} \left(RT\right)_{0}}\times p$ [By Eq. (i)]
$\Rightarrow \, \rho =\frac{3 \rho _{0}}{2}$