A diatomic gas of molecular mass 40 , g/mol i | vedclass

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(B) Let $n$ be the number of moles of the gas.The mass of the gas is $m = n 	imes M = n 	imes 40 \, g = 0.04n \, kg$.The kinetic energy of the gas m

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$\frac{320}{R} ^{\circ} C$

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(B) Let $n$ be the number of moles of the gas.The mass of the gas is $m = n 	imes M = n 	imes 40 \, g = 0.04n \, kg$.The kinetic energy of the gas moving with velocity $v = 200 \, m/s$ is $K.E. = \frac{1}{2} m v^2$.$K.E. = \frac{1}{2} 	imes (0.04n) 	imes (200)^2 = 0.02n 	imes 40000 = 800n \, J$.When the container is suddenly stopped,this kinetic energy is converted into internal energy of the gas,causing a rise in temperature $\Delta T$.The change in internal energy for a diatomic gas is $\Delta U = n C_v \Delta T$.For a diatomic gas,the degree of freedom $f = 5$,so $C_v = \frac{f}{2} R = \frac{5}{2} R$.Equating the kinetic energy to the change in internal energy: $800n = n 	imes \frac{5}{2} R 	imes \Delta T$.$\Delta T = \frac{800 	imes 2}{5R} = \frac{1600}{5R} = \frac{320}{R} ^{\circ} C$.

$A$ diatomic gas of molecular mass $40 \, g/mol$ is filled in a rigid container at temperature $30^{\circ} C$. It is moving with velocity $200 \, m/s$. If it is suddenly stopped,the rise in the temperature of the gas is .........

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