An ideal gas undergoes a process maintaining | vedclass

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

Question

(B) The given relation is $P = P_o(1 + \frac{V_o^2}{V^2})^{-1} = \frac{P_o V^2}{V^2 + V_o^2}$.From the ideal gas law,$PV = nRT$,so $T = \frac{PV}{nR}$

Vedclass · Accepted Answer

$\frac{11P_o V_o}{10R}$

Vedclass · Answer

(B) The given relation is $P = P_o(1 + \frac{V_o^2}{V^2})^{-1} = \frac{P_o V^2}{V^2 + V_o^2}$.From the ideal gas law,$PV = nRT$,so $T = \frac{PV}{nR}$.For sample $A$: $V_A = V_o$,so $P_A = \frac{P_o V_o^2}{V_o^2 + V_o^2} = \frac{P_o}{2}$.$T_A = \frac{P_A V_A}{nR} = \frac{(P_o/2) V_o}{2R} = \frac{P_o V_o}{4R}$.For sample $B$: $V_B = 3V_o$,so $P_B = \frac{P_o (3V_o)^2}{(3V_o)^2 + V_o^2} = \frac{9P_o V_o^2}{10V_o^2} = \frac{9P_o}{10}$.$T_B = \frac{P_B V_B}{nR} = \frac{(9P_o/10) (3V_o)}{2R} = \frac{27 P_o V_o}{20R}$.Now,the difference $T_B - T_A = \frac{27 P_o V_o}{20R} - \frac{P_o V_o}{4R} = \frac{27 P_o V_o}{20R} - \frac{5 P_o V_o}{20R} = \frac{22 P_o V_o}{20R} = \frac{11 P_o V_o}{10R}$.

Similar Questions

At the top of a mountain,a thermometer reads $7^{\circ}C$ and a barometer reads $70 \, cm$ of $Hg$. At the bottom of the mountain,these read $27^{\circ}C$ and $76 \, cm$ of $Hg$ respectively. The ratio of the density of air at the top to that at the bottom is

$A$ container holds $10 \ kg$ of gas at a pressure of $10^7 \ N/m^2$. How much gas in $kg$ must be removed so that the final pressure becomes $2.5 \times 10^6 \ N/m^2$? (Assume temperature remains constant.)

In the given pressure $(P)$ - absolute temperature $(T)$ graph of an ideal gas,the relation between volumes $V_1, V_2, V_3$ and $V_4$ is

What is molar volume? Write its $SI$ unit.

Vedclass Test Series

Exam Paper Generator

Online Exam Module