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The root mean square speed of oxygen molecules (O2) at a certain absolute temperature is v. If the temperature is doubled and oxygen gas dissociates into oxygen atom, the rms speed would be :

$v$

$\sqrt{2} v$

$2 v$

$2 \sqrt{2} v$

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Explanation

(c)

$V_{rms} = \sqrt{\frac{3 RT}{M}} V = \sqrt{\frac{3 RT}{M}} V' = \sqrt{\frac{3 R x 2 T}{M / 2}} = \sqrt{\frac{3 RT}{M}} x 2 V' = 2 v$

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If P is the pressure of the gas then the KE per unit volume of the gas is:

$\frac{P}{2}$

$P$

$\frac{3 P}{2}$

$2 P$

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Explanation

(c)

$P = \frac{1}{3} \frac{mn v^{2}}{V} \frac{(K . V .)}{V} = \frac{3 P}{2}$

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$n_{1}$ mole of monoatomic gas is mixed with $n_{2}$ mole of diatomic gas such that $γ_{m i x} = 1.5$

$n_{1} = 2 n_{2}$

$n_{2} = 2 n_{1}$

$n_{1} = n_{2}$

$n_{1} = 3 n_{2}$

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Explanation

(c)

For a mixture,

$γ_{m i x} = \frac{n_{1} C_{p_{1}} + n_{2} C_{p_{2}}}{n_{1} C_{v_{1}} + n_{2} C_{v_{2}}}$

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$Molar heat capacity of the process P = \frac{a}{T} for a monoatomic gas,' a' being positive constant i s -$

$\frac{7}{2} R$

$\frac{5}{2} R$

$\frac{3}{2} R$

None of these

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Explanation

$(1) PT = a P \frac{PV}{nR} = a P^{2} V = constant {PV}^{1 / 2} = constant {PV}^{x} = constant . . . . P o l y t r o p i c p r o c e s s w i t h x = \frac{1}{2} C = \frac{R}{γ - 1} + \frac{R}{1 - x} \frac{R}{(\frac{5}{3} - 1)} + \frac{R}{1 - \frac{1}{2}} = \frac{7}{2} R$

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Two chambers containin m₁g and m₂g of a gas at pressure P₁ and P₂ respectively are put in contact with each other. If temperature remains constant, the common pressure reached will be

$\frac{P_{1} P_{2} (m_{1} + m_{2})}{(P_{2} m_{1} + m_{2} P_{1})}$

$\frac{(m_{1} + m_{2})}{(P_{2} m_{1} + m_{2} P_{1})}$

$\frac{P_{1} P_{2}}{(P_{2} m_{1} + m_{2} P_{1})}$

$\frac{P_{1} P_{2}}{(m_{1} + m_{2})}$

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Explanation

$(1) According to Boyle' s law \frac{P}{ρ} = const . ∴ V_{1} = \frac{m_{1}}{ρ_{1}} = \frac{{km}_{1}}{P_{1}} and V_{2} = \frac{{km}_{2}}{P_{2}} Total volume = k (\frac{m_{1}}{P_{1}} + \frac{m_{2}}{P_{1}}) Let mixture has common pressure P and common density ρ ρ = \frac{(m_{1} + m_{2})}{k (\frac{m_{1}}{P_{1}} + \frac{m_{2}}{P_{2}})} \Rightarrow P = kρ = \frac{(m_{1} + m_{2})}{(\frac{m_{1}}{P_{1}} + \frac{m_{2}}{P_{2}})} = \frac{P_{1} P_{2} (m_{1} + m_{2})}{(P_{2} m_{1} + m_{2} P_{1})}$

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A rocket is propelled by a gas which is initially at a temperature of 4000 K. The temperature of the gas falls to 1000 K as it leaves the exhaust nozzle. The gas which will acquire the largest momentum while leaving the nozzle is:

Hydrogen

Helium

Nitrogen

Argon

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Explanation

Heavier gas will acquire largest momentum i.e. Argon.

$E_{g a s} = \frac{p_{g a s}^{2}}{2 M} M o r e t h e m o l e c u l a r m a s s M, m o r e i s t h e m o m e n t u m p_{g a s}^{} A r g o n h a s h i g h e s t M, s o l a r g e s t m o m e n t u m p_{g a s}^{}$

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A gas mixture consist of 2 moles of $O_{2}$ and 4 moles of Ar at temperature T. Neglecting all vibrational modes, the total internal energy of the system is:

4RT

15RT

9RT

11RT

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Explanation

(d)

The total internal energy of system =Internal energy of oxygen molecules+Internal

The energy of oxygen molecules+Internal energy of argon molecules

$= \frac{f_{1}}{2} n 1 R T + \frac{f_{2}}{2} n_{2} R T$

= $\frac{5}{2} \times 2 R T + \frac{3}{2} \times 4 R T$

$= 11 R T$

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One mole of an ideal monatomic gas undergoes a process described by the equation $P V^{3} =$ constant. The heat capacity of the gas during this process is:

$\frac{3}{2} R$

$\frac{5}{2} R$

$2 R$

$R$

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Explanation

(d)

As we know that for the polytropic process of index $α$ ,

Specific heat capacity= $C_{v} + \frac{R}{1 - α}$

$∵ P r o c e s s, P V^{3} = c o n s t a n t \Rightarrow α = 3$

$∴ C = C_{v} + \frac{R}{1 - α} = \frac{f R}{2} + \frac{R}{1 - 3}$

where, $C_{v} = \frac{f R}{2} = \frac{3 R}{2}$

For monatomic gas, $f = 3$

$\Rightarrow C = \frac{3 R}{2} - \frac{R}{2} = R$

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A given sample of an ideal gas occupies a volume V at a pressure p and absolute temperature T. The mass of each molecule of the gas is m. Which of the following gives the density of the gas?

p/(kT)

pm / (kT)

p/ (kTV)

mkT

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Explanation

(b)

As we know that

Pressure, $p = \frac{1}{3} . \frac{n m}{V} v_{r m s}^{2}$

$∵$ nm =mass of the gas, V= volume of the gas

$∴$ $\frac{m n}{V} =$ density of the gas. Thus,

$p = \frac{1}{2} ρ v_{r m s}^{2} = \frac{1}{3} ρ \frac{3 R T}{M_{0}} (∵ v_{r m s} = \sqrt{\frac{3 R T}{M_{0}}}) ρ = \frac{p M_{0}}{R T} = \frac{p m N_{A}}{k N_{A} T} [∵ R = N_{A} k a n d M_{0} = m N_{A}] ρ = \frac{p m}{k T}$

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The molecules of a given mass of gas have r.m.s velocity of 200 ms^-1 at 27°C and 1.0 x 10⁵ Nm^-2 pressure. When the temperature and pressure of the gas are respectively, 127°C and 0.05 X 10⁵Nm^-2 , the RMS velocity of its molecules in ms^-1 is:

400/√3

100√2/3

100/3

100√2

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Explanation

(a)

$I t i s g i v e n t h a t, v_{r m s} = 200 m s^{- 1}, T_{1} = 300 K, P_{1} = 10^{5} N / m^{2} T_{2} = 400 K, P_{2} = 0.05 X 10^{5} N / m^{2} v_{r m s} = \sqrt{3 R T / m} ∴ v_{r m s} \propto \sqrt T ∴ F o r t w o d i f f e r e n t c a s e s \Rightarrow \frac{v_{r m s_{1}}}{v_{r m s_{2}}} = \sqrt{\frac{T_{1}}{T_{2}}} = \sqrt{\frac{300}{400}} v_{r m s_{2}} = \frac{400}{\sqrt{3}} m / s$

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