Internal energy of Ideal gas

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Ideal gases $A$ , $B$ , and $C$ are contained in three different boxes which have exactly the same volume. The numbers of ideal gas molecules in box $A$ , $B$ , and $C$ are $6.02 \times 10^{23}$ , $12.04 \times 10^{23}$ , and $18.06 \times 10^{23},$ respectively. The temperatures of these gases are $200$ K, $300$ K, and $400$ K, respectively. Then what is the ratio of the internal energy of these gases?

A:B:C=2:3:4 A:B:C=1:3:6 A:B:C=1:2:3 A:B:C=6:8:9

1 solution

Lu Chee Ket
Jan 31, 2016

With $R = \frac{P_1 V_1}{n_1 T_1} = \frac{P_2 V_2}{n_2 T_2} = \frac{P_3 V_3}{n_3 T_3},$

$\frac{P_1}{1 \times 2} = \frac{P_2}{2 \times 3} = \frac{P_3}{3 \times 4}.$

$P_1 : P_2 : P_3 = 1 : 3: 6$

From $P = \frac13$ $p~\overline{c^2},$

Ratio = 1 : 3 : 6

Answer: $\boxed{A : B: C = 1 : 3 : 6}$

Internal energy of Ideal gas

1 solution

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