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An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity C remains constant. If during this process the relation of pressure P and volume V is given by PVⁿ=constant, then n is given by (Here C_P and C_V are molar specific heat at constant pressure and constant volume, respectively)
Option: 1   $n=\frac{C_{p}}{C_{v}}$

Option: 2   $n=\frac{C-C_{p}}{C-C_{v}}$

Option: 3 $n=\frac{C_{p}-C}{C-C_{v}}$

Option: 4   $n=\frac{C-C_{v}}{C-C_{p}}$

For a polytropic preocess

$c=c_{v}+\frac{R}{1-n} \: or \: \frac{R}{1-n} = c-c_{v}$

$\Rightarrow 1-n=\frac{R}{c-c_{v}} \: or\: n=1-\frac{R}{c-c_{v}}$

$\Rightarrow n=\frac{c-\left ( c_{v}+R \right )}{c-c_{v}} = \frac{c-c_{p}}{c-c_{v}}$

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Posted by

Ritika Jonwal

The temperature of an open room of volume 30 m³ increases from 17⁰C to 27⁰C due to the sunshine. The atmospheric pressure in the room remains 1×10⁵ Pa. If $n_{i}$ and $n_{f}$ are the number of molecules in the room before and after heating, then $n_{f}-n_{i}$ will be :
Option: 1 −1.61×10²³
Option: 2 1.38×10²³
Option: 3 2.5×10²⁵
Option: 4 −2.5×10²⁵

PV = nRT

$\Rightarrow\ \; n_{i}=\frac{PV}{RT_{i}},\ \; n_{f}=\frac{PV}{RT_{f}}$

$\Rightarrow\ \; n_{f}-n_{i}=\frac{PV}{R} \left(\frac{1}{T_{f}}-\frac{1}{T_{i}} \right )=\frac{10^{5}\times 30}{8.31}\times\left(\frac{1}{300}-\frac{1}{290} \right )$

$n_{f}-n_{i}=\frac{10^{5}\times 30}{8.31}\times\frac{-10}{300\times 290}=\frac{-10^{5}}{290\times 8.31}$

change in Number of molecules $=\frac{-10^{5}\times6.023\times10^{23}}{290\times 8.31}=-2.5\times 10^{25}$

Correct option is 4.

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vishal kumar

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The plot that depicts the behavior of the mean free time $\tau$ (time between two successive collisions) for the molecules of an ideal gas, as a function of temperature (T), qualitatively, is: (Graphs are schematic and not drawn to scale)
Option: 1

Option: 2

Option: 3

Option: 4

As relaxation time is given as

$\tau =\frac{\lambda}{V_{rms}}$

where λ=mean free path

and $V_{rms}\propto \sqrt{T}$

while $\boldsymbol{ \lambda=\frac{1}{\sqrt{2} \pi N d^{2}}}$

where N=number of molecules per unit volume

So keeping all the other quantities are constant.

we get $\tau \ \ \alpha \ \frac{1}{\sqrt{T}}$

So the correct graph is given in option 2

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Posted by

vishal kumar

Two moles of an ideal gas with $\frac{C_p}{C_v}=\frac{5}{3}$ are mixed 3 moles of another ideal gas with $\frac{C_p}{C_v}=\frac{4}{3}$ . The value of $\frac{C_p}{C_v}$ for the mixture is:-
Option: 1 $1.50$
Option: 2 $1.45$
Option: 3 $1.47$
Option: 4 $1.42$

For ideal gas:- $C_p-C_v=R$

For first case:-

$\frac{C_{p1}}{C_{v1}}=\frac{5}{3} \ and \ C_{p1}-C_{v1}=R$

$C_{p1}=\frac{5}{3}{C_{v1}} \ and \ \frac{5}{3}{C_{v1}} -C_{v1}=R\Rightarrow \frac{2}{3}C_{v1}=R\Rightarrow C_{v1}=\frac{3}{2}R$

So, $C_{p1}=\frac{5}{2}R$

For second case:-

$\frac{C_{p2}}{C_{v2}}=\frac{4}{3} \ and \ C_{p2}-C_{v2}=R$

$C_{p2}=\frac{4}{3}{C_{v2}} \ and \ \frac{4}{3}{C_{v2}}-C_{v2}=R\Rightarrow C_{v2}=3R$ and $C_{p2}=4R$

Now, $Y_{mix}=\frac{n_1C_{p1}+n_2C_{p2}}{n_1C_{v1}+n_2C_{v2}}=\frac{2\times\frac{5}{2}R+3\times4R}{2\times\frac{3}{2}R+3\times3R}= 1.417 = 1.42$

So option (4) is correct.

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Ritika Jonwal

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Two gases-argon (atomic radius 0.07 nm, atomic weight 40) and xenon ( atomic radius 0.1 nm, atomic weight 140) have the same number density and are at the same temperature. The ratio of their respective mean free times is closest to :
Option: 1 $4.67$
Option: 2 $2.3$
Option: 3 $1.83$
Option: 4 1.07

Mean free time $=\frac{1}{\sqrt{2n}_{n}d^{2}v_{ms}}$

$\frac{t_{Ar}}{t_{xe}}=\frac{d^{2}_{xe}}{d^{2}_{Ar}}\times \sqrt{\frac{m_{1}}{m_{2}}}$

$=\left ( \frac{0.1}{0.07} \right )^{2}\times \sqrt{\frac{40}{140}}$

$=\left ( \frac{10}{7} \right )^{2}\times \sqrt{\frac{2}{7}}=1.07$

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avinash.dongre

Consider a mixture of n moles of helium gas and 2n moles of oxygen gas (molecules taken to be rigid) as an ideal gas. Its C_p/C_v value will be :
Option: 1 40/27
Option: 2 23/15
Option: 3 19/13
Option: 4 67/45

For monoatomic gas (He)

$c_v=\frac{3R}{2}, c_p=\frac{5R}{2}$

For diatomic gas (O₂)

$c_v=\frac{5R}{2}, c_p=\frac{7R}{2}$

$\\\gamma_{\text {mixure }}=\frac{C_{p_{m i x}}}{C_{v_{m x}}}=\frac{\left(n_{1} C_{p_{1}}+n_{2} C_{p_{2}}\right)}{\left(n_{1} C_{v_{1}}+n_{2} C_{v_{2}}\right)}\\\\=\frac{n\frac{5}{2}R+2n\frac{7}{2}R}{n\frac{3}{2}R+2n\frac{5}{2}R}=\frac{19}{13}$

Hence the correct option is (3).

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Posted by

vishal kumar

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Under an adiabatic process, the volume of an ideal gas gets doubled. Consequently the mean collision time between the gas molecule changes from $\tau_{1}$ to $\tau_{2}$ . If $\frac{C_{p}}{C_{v}}=\gamma$ for this gas then a good estimate for $\frac{\tau_{2}}{\tau_{1}}$ is
Option: 1 $\left(\frac{1}{2}\right)^{(\gamma+1)/{2}}$

Option: 2 $\frac{1}{2}$

Option: 3 $\left ( \frac{1}{2} \right )^{\gamma }$

Option: 4 $(2)^{{(1+\gamma})/{2}}$

As, relaxation time= $\tau \ \ \alpha \ \ \frac{V}{\sqrt{T}}$

and using $PV^{\gamma }=Const$ gives $\mathrm{T} \propto \frac{1}{\mathrm{V}^{\gamma-1}}$

$\begin{array}{l}{\tau \propto V^{1+{(\gamma-1)}/{2}}} \\ {\tau \propto V^{{(1+\gamma)}/{2}}}\end{array}$

$\begin{aligned} &\frac{\tau_{2}}{\tau_{\mathrm{1}}}=\left(\frac{2 \mathrm{~V}}{\mathrm{~V}}\right)^{{(1+\gamma})/{2}} \\ &\frac{\tau_{\mathrm{2}}}{\tau_{\mathrm{1}}}=(2)^{(1+\gamma) / 2} \end{aligned}$

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Ritika Jonwal

Consider two ideal diatomic gases A and B at same temperature T. Molecules of the gas A are rigid and have a mass m. Molecules of the gas B have an additional vibrational mode and have a mass $\frac{m}{4}$ . The ratio of the specific heats $\left(C^A_V \textup{ and } C_V^B \right )$ of gas A and B respectively is:
Option: 1 5:9
Option: 2 7:9
Option: 3 3:5
Option: 4 5:7

Molar heat capacity of A at constant volume $=\frac{5R}{2}$

Molar heat capacity of B at constant volume $=\frac{7R}{2}$

Dividing both $\frac{(C_V)_A}{C_V)_B}=\frac{5}{7}$

Hence the correct option is (4).

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avinash.dongre

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The RMS speeds of the molecules of Hydrogen , Oxygen and Carbon dioxide at the same temperature are $V_{H},V_{O}\, and\, V_{C}$ respectively then:
Option: 1 $V_{C}> V_{O}>V_{H}$
Option: 2 $V_{H}= V_{O}> V_{C}$
Option: 3 $V_{H}> V_{O}> V_{C}$
Option: 4 $V_{H}= V_{O}= V_{C}$

$V_{rms}= \sqrt{\frac{3RT}{M}}$
$M_{Hydrogen}= 2\, amu$
$M_{Oxygen}= 32\, amu$
$M_{CO_{2}}= 44\, amu$
$V_{H}> V_{O}> V_{C}$
The correct option is (3)

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Posted by

vishal kumar

The temperature of equal masses of three ifferent liquids x,y and z are $10^{\circ}C,20^{\circ}C \, and\, 30^{\circ}C$ respectively.The temperature of mixture when x is mixed with y is $16^{\circ}C$ and that when y is mixed with z is $26^{\circ}C$ , The temperature of mixture when x and z are mixed will be :

Option: 1 $25\cdot 62^{\circ}C$
Option: 2 $20\cdot 28^{\circ}C$
Option: 3 $28\cdot 32^{\circ}C$
Option: 4 $23\cdot 84^{\circ}C$

$T_{mix}= \frac{n_{1}T_{1}+n_{2}T_{2}}{n_{1}+n_{2}}$
$T_{x}= 10^{\circ}C= 283\, k$
$T_{y}= 20^{\circ}C= 293\, k$
$T_{z}= 30^{\circ}C= 303\, k$
For x & y mixture ,
$T_{1}= \frac{n_{x}T_{x}+n_{y}T_{y}}{n_{x}+n_{y}}$
$289= \frac{\frac{m_{x}T_{x}}{M_{x}}+\frac{m_{y}T_{y}}{M_{y}}}{\frac{m_{x}}{M_{x}}+\frac{m_{y}}{M_{y}}}$
$289= \frac{\frac{283}{M_{x}}+\frac{293}{M_{y}}}{\frac{1}{M_{x}}+\frac{1}{M_y}}\rightarrow \left ( 1 \right )$
Mixture of y & Z
$299= \frac{\frac{293}{M_{y}}+\frac{303}{M_{z}}}{\frac{1}{M_{y}}+\frac{1}{M_z}}\rightarrow \left ( 2 \right )$
Mixture of x & z
$T_{3}= \frac{\frac{283}{M_{x}}+\frac{303}{M_{z}}}{\frac{1}{M_{x}}+\frac{1}{M_z}}\rightarrow \left ( 3 \right )$
From eqn (1)
$\frac{289}{M_{x}}+\frac{289}{M_{y}}= \frac{283}{M_{x}}+\frac{213}{M_{y}}$
$6M_{y} = 4M_{x}$
$3M_{y} = 2M_{x}\rightarrow \left ( 4 \right )$
$6M_{z} = 4M_{y}$
$3M_{z} = 2M_{y}\rightarrow \left ( 5 \right )$
$3\times \frac{3M_{z}}{2} = 2M_{x}$
$9 M_{z} = 4M_{x}\rightarrow \left (6 \right )$
$T_{3}= \frac{\frac{283}{\frac{9}{4}M_{z}}+\frac{303}{M_{z}}}{\frac{1}{\frac{9}{4}M_{z}}+\frac{1}{M_{z}}}= \frac{\frac{4\times 283}{9}+\frac{303\times 9}{9}}{\frac{4}{9}+\frac{9}{9}}$
$= \frac{4\times283+303\times9}{13}= 296\cdot 84 k$
$T_{3}= 23\cdot 84\degree C$
The correct option is (4)