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One mole of ideal monoatomic gas $ \left( \gamma = {5 \over 3 } \right) $ is mixed with one mole of diatomic gas $ \left( \gamma = {7 \over 3 } \right) $ What is $\gamma $ for the mixture? $\gamma $ denotes the ratio of specific heat at constant pressure to that at constant volume

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Explanation

$ \gamma _{mix} = { { 1\gamma_1 \over \gamma_1 -1 } + { 2\gamma_2 \over \gamma_2 -1 } \over { 1 \over \gamma_1 - 1 } + {2 \over \gamma_2 -1 } } = { { 1 \times 5/3 \over ( 5/3 - 1 ) } + { 1 \times 7/5 \over 7/5 - 1 } \over {1 \over (5/3 -1)} + {1 \over ( 7/5 -1 ) } }= { 3 \over 2} = 1.5 $

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For a gas if ratio of specific heats at constant pressure and volume is $ \gamma $, then value of degrees of freedom is

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Explanation

$ \gamma = 1 + { 2 \over f } \Rightarrow \gamma -1 = { 2 \over f } \Rightarrow { f \over 2} = { 1 \over \gamma -1 } $ $ \therefore f = { 2 \over \gamma - 1 } $

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The molar specific heat at constant pressure of an ideal gas is The ratio of specific heat at constant pressure to that ratio at constant volume is

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Explanation

molar specific heat at constant pressure, $C_p = 7 \over 2 R $ Since $ C_p - C_v = R \Rightarrow C_v = C_p - R = { 7 \over 2} R -R = { 5 \over 2 } R $

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For a gas $ \left( \gamma = {7 \over 5 } \right) $ , the gas may probably be

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Explanation

$ \gamma = 7/5 $ for a diatomic gas

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Direction:- Assertion : 300 cc of a gas at $ 27^\circ C$ is cooled at $ -3 ^\circ C$ at constant pressure. The final volume of the gas would be 270 cc Reason : This is as per charle's law $ { V_2 \over V_1 } = { T_2 \over T_1} $

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Direction:- Assertion : The time of collision of molecules is of the order of $10^{-8} s $, which is very very small compared to the time between two successive collisions. Reason : This is an experimental fact.

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Explanation

The time of collision of molecules being of the order of \( 10^{-8} \text{ s} \\) is an experimental fact, and it is true. This time is very small compared to the time between two successive collisions. Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

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Direction:- Assertion : Mean free path of gas varies inversly as density of the gas. Reason : Mean free path varies inversely as pressure of the gas.

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Explanation

The mean free path (\

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The velocity of a body of mass 20 kg decreases from $20 ms^{–1} to 5 ms^{–1}$ in a distance of 100 m. Force on the body is

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Explanation

$ F = ma = m \left( \nu^2 - u^2 \over 2S \right) $

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A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2 m while applying the force and the ball goes upto 2 m height further, find the magnitude of the force.$ (Consider g = 10 ms^{–2}) $

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Explanation

$ H_{max} = { u^2 \over 2g } $ $ \therefore u = \sqrt { 2gH_{max} }$ This velocity is supplied to the ball by hand and initially the hand was at rest. It acquires this velocity in distance of 0.2 meter. $ \therefore a = { u^2 \over 2S } $ So upword force F = m (g+a)

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Formula for true force is

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Explanation

According to newton's second law force = rate of change of linear momentum.

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