The density $(\rho)$ versus pressure $(P)$ of a given mass of an ideal gas is shown at two temperatures $T_{1}$ and $T_{2}$ [image:13_image3.emf] Then relation between $T_{1}$ and $T_{2}$ may be

Question

Dr. Abubacker · Accepted Answer

(b) According to ideal gas equation $PV = nRT$
$PV = \frac{m}{M}RT,\ P = \frac{\rho}{M}RT$ or $\frac{\rho}{P} = \frac{M}{RT}$or $\frac{\rho}{P} \propto \frac{1}{T}$
Here, $\frac{\rho}{P}$ represent the slope of graph
Hence $T_{2} > T_{1}$

Q. The density $(\rho)$ versus pressure $(P)$ of a given mass of an ideal gas is shown at two temperatures $T_{1}$ and $T_{2}$

Then relation between $T_{1}$ and $T_{2}$ may be

Explanation

Reviewed by Dr. Abubacker

Want more practice?

Q. The density $(\rho)$ versus pressure $(P)$ of a given mass of an ideal gas is shown at two temperatures $T_{1}$ and $T_{2}$ Then relation between $T_{1}$ and $T_{2}$ may be

Explanation

Reviewed by Dr. Abubacker

Want more practice?

Q. The density $(\rho)$ versus pressure $(P)$ of a given mass of an ideal gas is shown at two temperatures $T_{1}$ and $T_{2}$

Then relation between $T_{1}$ and $T_{2}$ may be