One mole of an ideal gas passes through a pro | vedclass

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

Question

(C) Given $n = 1$ mole.From the ideal gas equation,$PV = nRT = RT$,so $P = \frac{RT}{V}$.Substituting this into the given relation $P = P_0 \left[ 1 -

Vedclass · Accepted Answer

$\frac{5}{4} \frac{P_0 V_0}{R}$

Vedclass · Answer

(C) Given $n = 1$ mole.From the ideal gas equation,$PV = nRT = RT$,so $P = \frac{RT}{V}$.Substituting this into the given relation $P = P_0 \left[ 1 - \frac{1}{2} \left( \frac{V_0}{V} \right)^2 \right]$:$\frac{RT}{V} = P_0 \left[ 1 - \frac{V_0^2}{2V^2} \right]$$T = \frac{P_0}{R} \left[ V - \frac{V_0^2}{2V} \right]$Now,calculate temperature at $V = V_0$ and $V = 2V_0$:$T_1 = T(V_0) = \frac{P_0}{R} \left[ V_0 - \frac{V_0^2}{2V_0} \right] = \frac{P_0}{R} \left[ V_0 - \frac{V_0}{2} \right] = \frac{P_0 V_0}{2R}$$T_2 = T(2V_0) = \frac{P_0}{R} \left[ 2V_0 - \frac{V_0^2}{2(2V_0)} \right] = \frac{P_0}{R} \left[ 2V_0 - \frac{V_0}{4} \right] = \frac{P_0}{R} \left[ \frac{7V_0}{4} \right] = \frac{7 P_0 V_0}{4R}$Change in temperature $\Delta T = T_2 - T_1 = \frac{7 P_0 V_0}{4R} - \frac{P_0 V_0}{2R} = \frac{7 P_0 V_0 - 2 P_0 V_0}{4R} = \frac{5 P_0 V_0}{4R}$.

$A$. Process $A$	$P=5, V=7$
$B$. Process $B$	$P=4, V=6$
$C$. Process $C$	$P=12, V=1$
$D$. Process $D$	$P=6, V=3$

One mole of an ideal gas passes through a process where pressure and volume obey the relation $P = P_0 \left[ 1 - \frac{1}{2} \left( \frac{V_0}{V} \right)^2 \right]$. Here $P_0$ and $V_0$ are constants. Calculate the change in the temperature of the gas if its volume changes from $V_0$ to $2V_0$.

Similar Questions

Air is filled at $60^{\circ} C$ in a vessel of open mouth. The vessel is heated to a temperature $t^{\circ} C$ so that $\frac{1}{4}$th of the air escapes from the vessel. Assuming air as an ideal gas and the volume of the vessel remains constant, the value of '$t$' is: (in $^{\circ} C$)

An ideal gas at atmospheric pressure has a temperature of $300 \,K$ and a volume of $1 \,m^3$. If its temperature and volume are both doubled,its pressure will be ...........

An ideal gas initially at pressure $1 \, bar$ is being compressed from $30 \, m^{3}$ to $10 \, m^{3}$ volume and its temperature decreases from $320 \, K$ to $280 \, K$. Find the final pressure of the gas (in $bar$).

$Assertion :$ One mole of any substance at any temperature or volume always contains $6.02 \times 10^{23}$ molecules.
$Reason :$ One mole of a substance always refers to $S.T.P.$ conditions.

Vedclass Test Series

Exam Paper Generator

Online Exam Module