If 2 text{ moles} of an ideal monoatomic gas | vedclass

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(B) For an ideal monoatomic gas,the internal energy is given by $U = n C_v T$,where $C_v = \frac{3}{2} R$.Since the gases are mixed in an isolated sys

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$227$

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(B) For an ideal monoatomic gas,the internal energy is given by $U = n C_v T$,where $C_v = \frac{3}{2} R$.Since the gases are mixed in an isolated system,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 $C_v$ is the same for both monoatomic gases,it cancels out:$(n_1 + n_2) T_{mix} = n_1 T_1 + n_2 T_2$.Given: $n_1 = 2 	ext{ mol}$,$T_1 = 27 + 273 = 300 	ext{ K}$,$n_2 = 4 	ext{ mol}$,$T_2 = 327 + 273 = 600 	ext{ K}$.$(2 + 4) T_{mix} = (2 	imes 300) + (4 	imes 600)$.$6 T_{mix} = 600 + 2400 = 3000$.$T_{mix} = \frac{3000}{6} = 500 	ext{ K}$.Converting back to Celsius: $T_{mix} = 500 - 273 = 227^{\circ} C$.

If $2 \text{ moles}$ of an ideal monoatomic gas at a temperature of $27^{\circ} C$ is mixed with $4 \text{ moles}$ of another ideal monoatomic gas at a temperature of $327^{\circ} C$,then the temperature of the mixture of the two gases is: (in $^{\circ} C$)

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