One mole of an ideal monoatomic gas (gamma = | vedclass

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

Question

(A) For a mixture of gases,the adiabatic index $\gamma_{	ext{mix}}$ is given by the formula:$\gamma_{	ext{mix}} = \frac{n_1 C_{p,1} + n_2 C_{p,2}}{n

Vedclass · Accepted Answer

$3/2$

Vedclass · Answer

(A) For a mixture of gases,the adiabatic index $\gamma_{	ext{mix}}$ is given by the formula:$\gamma_{	ext{mix}} = \frac{n_1 C_{p,1} + n_2 C_{p,2}}{n_1 C_{v,1} + n_2 C_{v,2}}$Alternatively,using the relation $C_v = \frac{R}{\gamma - 1}$,we have:$\gamma_{	ext{mix}} = \frac{n_1 + n_2}{\frac{n_1}{\gamma_1 - 1} + \frac{n_2}{\gamma_2 - 1}}$Given $n_1 = 1, \gamma_1 = 5/3$ and $n_2 = 1, \gamma_2 = 7/5$:$\gamma_{	ext{mix}} = \frac{1 + 1}{\frac{1}{5/3 - 1} + \frac{1}{7/5 - 1}} = \frac{2}{\frac{1}{2/3} + \frac{1}{2/5}} = \frac{2}{3/2 + 5/2} = \frac{2}{8/2} = \frac{2}{4} = 1/2$ (Wait,re-calculating).Correct calculation:$C_{v,1} = \frac{R}{5/3 - 1} = \frac{3R}{2}$,$C_{v,2} = \frac{R}{7/5 - 1} = \frac{5R}{2}$.$C_{v,	ext{mix}} = \frac{n_1 C_{v,1} + n_2 C_{v,2}}{n_1 + n_2} = \frac{1(1.5R) + 1(2.5R)}{2} = 2R$.$C_{p,	ext{mix}} = C_{v,	ext{mix}} + R = 2R + R = 3R$.$\gamma_{	ext{mix}} = \frac{C_{p,	ext{mix}}}{C_{v,	ext{mix}}} = \frac{3R}{2R} = 1.5 = 3/2$.

One mole of an ideal monoatomic gas $(\gamma = 5/3)$ is mixed with one mole of a diatomic gas $(\gamma = 7/5).$ What is $\gamma$ for the mixture? $\gamma$ denotes the ratio of specific heat at constant pressure to that at constant volume.

Similar Questions

$A$ gas mixture consists of molecules of type $1, 2$ and $3$,with molar masses $m_1 > m_2 > m_3$. $v_{rms}$ and $\bar K$ are the $r.m.s.$ speed and average kinetic energy of the gases. Which of the following is true?

Three moles of oxygen are mixed with $2 \, moles$ of helium. What will be the ratio of specific heats at constant pressure and constant volume for the mixture?

$A$ rigid tank contains $35 \, kg$ of nitrogen at $6 \, atm$. $A$ sufficient quantity of oxygen is supplied to increase the pressure to $9 \, atm$,while the temperature remains constant. The amount of oxygen supplied to the tank is .... $kg$.

$A$ gaseous mixture has $2$ moles of oxygen and $4$ moles of Argon at a temperature $T$. Neglecting all vibrational modes of the molecules,the total internal energy of the system is ($R$ = Universal gas constant). (in $RT$)

The temperature of the mixture of one mole of helium and one mole of hydrogen is increased from $0^{\circ}C$ to $100^{\circ}C$ at constant pressure. The amount of heat delivered will be ...... $cal$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module