The pressure-temperature graph for an ideal g | vedclass

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

Question

(B) From the ideal gas equation,$PV = nRT$,where $n = \frac{m}{M}$.So,$PV = \frac{m}{M} RT$,which gives $P = \frac{m}{V} \frac{RT}{M} = \rho \frac{RT}

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) From the ideal gas equation,$PV = nRT$,where $n = \frac{m}{M}$.So,$PV = \frac{m}{M} RT$,which gives $P = \frac{m}{V} \frac{RT}{M} = \rho \frac{RT}{M}$.This implies $P \propto \rho T$,or $\frac{P_1}{P_2} = \frac{\rho_1 T_1}{\rho_2 T_2}$.Given at point $A$: $P_A = P_0$,$T_A = T_0$,$\rho_A = \rho_0$.Given at point $B$: $P_B = 3P_0$,$T_B = 2T_0$,$\rho_B = ?$.Substituting the values: $\frac{P_0}{3P_0} = \frac{\rho_0 T_0}{\rho_B (2T_0)}$.$\frac{1}{3} = \frac{\rho_0}{2 \rho_B} \Rightarrow \rho_B = \frac{3}{2} \rho_0$.

The pressure-temperature graph for an ideal gas is shown in the figure. At point $A$,the density of the gas is $\rho_0$. What will be the density at point $B$?

Similar Questions

The number of molecules in a gas at pressure $1.64 \times 10^{-3} \text{ atm}$ and temperature $200 \text{ K}$ having the volume $1 \text{ cc}$ is:

In the equation $PV = KT$,the value of $K$ depends on which factors?

The pressure $Vs$ temperature graph of an ideal gas is shown in the figure. If the density of the gas at point $A$ is $\rho _0$,what will be the density at point $B$?

At constant pressure,the ratio of the increase in volume of an ideal gas per degree rise in Kelvin temperature to its original volume is $(T =$ absolute temperature of the gas$)$

If the volume of a given mass of a gas is increased four times,and the temperature is raised from $27^{\circ}C$ to $127^{\circ}C$,what happens to its elasticity?

Vedclass Test Series

Exam Paper Generator

Online Exam Module