Question 694: Thermodynamics & KTG

Question

For an ideal gas, the density of the gas is $\rho_0$ when temperature and pressure of the gas are $\mathrm{T}_0$ and $P_0$ respectively. when the temperature of the gas is $2 \mathrm{~T}_0$, its pressure becomes $3 \mathrm{P}_0$. The new density will be

Accepted Answer

Let the mass of the gas be $m$, and its volume at initial conditions be $V_0$.
Initial condition:
Pressure $= P_0$
Temperature $= T_0$
Density $= \rho_0$
We know,
$$ \rho_0 = \frac{m}{V_0} $$
Using the ideal gas equation,
$$ P_0 V_0 = n R T_0 $$
Where $n$ is the number of moles.
$n = \frac{m}{M}$ (where $M$ is the molar mass)
So,
$$ P_0 V_0 = \frac{m}{M} R T_0 $$
Final condition:
Pressure $= 3P_0$
Temperature $= 2T_0$
Volume $= V$
Again from the ideal gas law,
$$ 3P_0 \cdot V = \frac{m}{M} R \cdot 2T_0 $$
Solving for $V$ in final condition:
$$ 3P_0 V = \frac{m}{M} R \cdot 2T_0 $$
But from the initial condition,
$$ P_0 V_0 = \frac{m}{M} R T_0 \implies \frac{m}{M} R T_0 = P_0 V_0 $$
So, substitute:
$$ 3P_0 V = 2 P_0 V_0 \implies V = \frac{2}{3} V_0 $$
Find new density:
$$ \rho = \frac{m}{V} = \frac{m}{(2/3)V_0} = \frac{3m}{2V_0} $$
But $\rho_0 = \frac{m}{V_0}$
So,
$$ \rho = \frac{3}{2} \rho_0 $$
Final Answer:
The new density is
$ \boxed{\frac{3}{2}\rho_0} $
Correct Option: D

Answer

$\frac{2}{3} \rho_0$

Answer

$\frac{3}{4} \rho_0$

Answer

$\frac{4}{3} \rho_0$

Answer

$\frac{3}{2} \rho_0$

Thermodynamics & KTG

For an ideal gas, the density of the gas is \rho_0 when temperature and pressure of the gas are \mathrm{T}_0 and P_0 respectively. when the temperature of the gas is 2 \mathrm{~T}_0, its pressure becomes 3 \mathrm{P}_0. The new density will be

Choose the correct answer: