The following graph shows two isotherms for a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) For an ideal gas,the equation of state is $PV = \mu RT$. Since the mass is fixed,$\mu$ is constant. Thus,$T \propto PV$.From the graph,we can choo

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) For an ideal gas,the equation of state is $PV = \mu RT$. Since the mass is fixed,$\mu$ is constant. Thus,$T \propto PV$.From the graph,we can choose points on the isotherms to find the product $PV$ for each temperature.For isotherm $T_1$,at $V = 1 	ext{ m}^3$,$P = 1 	imes 10^5 	ext{ Pa}$. So,$T_1 \propto (1 	imes 1) = 1$.For isotherm $T_2$,at $V = 2 	ext{ m}^3$,$P = 2 	imes 10^5 	ext{ Pa}$. So,$T_2 \propto (2 	imes 2) = 4$.Therefore,the ratio of temperatures is $\frac{T_2}{T_1} = \frac{4}{1} = 4$.The $r.m.s.$ speed of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$,which implies $v_{rms} \propto \sqrt{T}$.Thus,the ratio of $r.m.s.$ speeds is $\frac{v_{rms,2}}{v_{rms,1}} = \sqrt{\frac{T_2}{T_1}} = \sqrt{4} = 2$.

The following graph shows two isotherms for a fixed mass of an ideal gas. The ratio of $r.m.s.$ speed of the molecules at temperatures $T_2$ and $T_1$ is

Similar Questions

The $r.m.s.$ speed of the molecules of a gas in a vessel is $400 \; m/s$. If half of the gas leaks out,at constant temperature,the $r.m.s.$ speed of the remaining molecules will be ..... $m/s$.

At $300\,K$,the rms speed of oxygen molecules is $\sqrt{\frac{\alpha+5}{\alpha}}$ times its average speed in the gas. Then,the value of $\alpha$ will be (use $\pi=\frac{22}{7}$):

If the absolute temperature of a gas is increased $5$ times,the r.m.s. velocity of the gas molecules will be

For a gas at a temperature $T$,the root-mean-square velocity ${v_{rms}}$,the most probable speed ${v_{mp}}$,and the average speed ${v_{av}}$ obey the relationship:

For a gas at room temperature of $27^{\circ}C$,the value of $\nu_{rms}$ is $1930 \, m/s$. The gas is.......... $(R = 8.3 \, J/mol \cdot K)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module