If 4 moles of an ideal monoatomic gas at temp | vedclass

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(D) For an ideal monoatomic gas,the internal energy is given by $U = n C_V T$,where $C_V = \frac{3}{2} R$.When two gases are mixed in an adiabatic con

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$500 ~K$

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(D) For an ideal monoatomic gas,the internal energy is given by $U = n C_V T$,where $C_V = \frac{3}{2} R$.When two gases are mixed in an adiabatic container,the total internal energy is conserved: $U_{mix} = U_1 + U_2$.$(n_1 + n_2) C_V T_{mix} = n_1 C_V T_1 + n_2 C_V T_2$.Since the gases are both monoatomic,$C_V$ cancels out:$T_{mix} = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2}$.Substituting the given values:$T_{mix} = \frac{4(400) + 2(700)}{4 + 2}$.$T_{mix} = \frac{1600 + 1400}{6} = \frac{3000}{6} = 500 ~K$.

If $4$ moles of an ideal monoatomic gas at temperature $400 ~K$ is mixed with $2$ moles of another ideal monoatomic gas at temperature $700 ~K$,the temperature of the mixture is

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