One mole of ideal monoatomic gas (gamma = 5/3 | vedclass

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

Question

(A) For a monoatomic gas,$C_{v1} = \frac{3}{2}R$ and $C_{p1} = \frac{5}{2}R$. For a diatomic gas,$C_{v2} = \frac{5}{2}R$ and $C_{p2} = \frac{7}{2}R$.G

Vedclass · Accepted Answer

$3/2$

Vedclass · Answer

(A) For a monoatomic gas,$C_{v1} = \frac{3}{2}R$ and $C_{p1} = \frac{5}{2}R$. For a diatomic gas,$C_{v2} = \frac{5}{2}R$ and $C_{p2} = \frac{7}{2}R$.Given $n_1 = 1$ mole and $n_2 = 1$ mole.The equivalent molar heat capacity at constant volume for the mixture is $(C_v)_{	ext{mix}} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} = \frac{1(\frac{3}{2}R) + 1(\frac{5}{2}R)}{1 + 1} = \frac{4R}{2} = 2R$.The equivalent molar heat capacity at constant pressure for the mixture is $(C_p)_{	ext{mix}} = \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1 + n_2} = \frac{1(\frac{5}{2}R) + 1(\frac{7}{2}R)}{1 + 1} = \frac{6R}{2} = 3R$.The adiabatic index for the mixture is $\gamma_{	ext{mix}} = \frac{(C_p)_{	ext{mix}}}{(C_v)_{	ext{mix}}} = \frac{3R}{2R} = 1.5 = 3/2$.

One mole of ideal monoatomic gas $(\gamma = 5/3)$ is mixed with one mole of 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

$22 \ g$ of carbon dioxide at $27^{\circ} C$ is mixed in a closed container with $16 \ g$ of oxygen at $37^{\circ} C$. If both gases are considered as ideal gases,then the temperature of the mixture is nearly: (in $^{\circ} C$)

$1 \, g$ of $H_2$ at $27 \, ^oC$ is mixed with $16 \, g$ of $O_2$ at $37 \, ^oC$. The temperature of the mixture is about ....... $^oC$.

The molar heat capacity of a mixture of two gases at constant volume is $13R/6$. The ratio of the number of moles of the first gas to the second is $1:2$. The respective gases may be:

Two gases $A$ and $B$ having same pressure $P$,volume $V$,and absolute temperature $T$ are mixed. If the mixture has the volume and temperature as $V$ and $T$ respectively,then the pressure of the mixture is

The volume $V$ of an enclosure contains a mixture of three gases: $16 \, g$ of oxygen,$28 \, g$ of nitrogen,and $44 \, g$ of carbon dioxide at absolute temperature $T$. Consider $R$ as the universal gas constant. The pressure of the mixture of gases is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module