Question 781: Thermodynamics & KTG

Question

For an ideal gas the density of the gas is $$\rho_0$$ when temperature and pressure of the gas are $$T_0$$ and $$P_0$$ respectively. When the temperature of the gas is $$2 \mathrm{~T}_0$$, its pressure will be $$3 \mathrm{P}_0$$. The new density will be

Accepted Answer

Density $$\propto \mathrm{P} / \mathrm{T}$$
So, $$\frac{\mathrm{d}_2}{\mathrm{~d}_1}=\frac{\mathrm{P}_2}{\mathrm{~T}_2} 	imes \frac{\mathrm{T}_1}{\mathrm{P}_1}$$
$$	herefore \quad$$ The new density is:
$$\begin{aligned}
& \mathrm{d}_2=ho_0 	imes \frac{3 \mathrm{P}_0}{2 \mathrm{~T}_0} 	imes \frac{\mathrm{T}_0}{\mathrm{P}_0} \
& \mathrm{~d}_2=\frac{3}{2} ho_0
\end{aligned}$$

Answer

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

Answer

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

Answer

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

Answer

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

Thermodynamics & KTG

For an ideal gas the density of the gas is $\rho_0 when temperature and pressure of the gas are T_0 and P_0 respectively. When the temperature of the gas is 2 \mathrm{~T}_0, its pressure will be 3 \mathrm{P}_0$. The new density will be

Choose the correct answer: