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(D) The ideal gas equation is given by $PV = nRT$. Since $n = \frac{m}{M}$,where $m$ is the mass and $M$ is the molar mass,we have $PV = \frac{m}{M} R

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$\frac{3}{2} \varrho_0$

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(D) The ideal gas equation is given by $PV = nRT$. Since $n = \frac{m}{M}$,where $m$ is the mass and $M$ is the molar mass,we have $PV = \frac{m}{M} RT$. Rearranging this,we get $P = \frac{m}{V} \frac{RT}{M} = \varrho \frac{RT}{M}$,where $\varrho$ is the density. Thus,$\varrho = \frac{PM}{RT}$. For the initial state: $\varrho_0 = \frac{P_0 M}{R T_0}$. For the final state: $\varrho' = \frac{P' M}{R T'} = \frac{(3 P_0) M}{R (2 T_0)}$. Substituting $\varrho_0$ into the expression for $\varrho'$,we get $\varrho' = \frac{3}{2} \left( \frac{P_0 M}{R T_0} \right) = \frac{3}{2} \varrho_0$.

For an ideal gas,the density of the gas is $\varrho_0$ when temperature and pressure of the gas are $T_0$ and $P_0$ respectively. When the temperature of the gas is $2 T_0$,its pressure becomes $3 P_0$. The new density will be

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