# If the speed of sound in Hydrogen at NTP is equal to the speed of sound in carbon-di-oxide at $T_0$, what is the value of $T_0$? (i.e, $T = 273K$)

(A) 2.4 T (B) 24 T (C) .24 T (D) 0.024 T

## 1 Answer

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• Molar mass of $CO_2$ is 44.00964 ± 0.00003 g/mol and Molar mass of $H_2$ is 2.015894 ± 0.000002 g/mol
• $v_{sound} = \large \sqrt \frac{\gamma k_BT}{M}$, where $k_B$ is the Boltzman's constant and M is the molar mass of the gas, $\gamma$ is the adiabatic index and T is the temperature.
• $\gamma$ for $H_2 = \large\frac{7}{5}$
• $\gamma$ for $CO_2 = 1.28$
$v_{sound} = \large \sqrt \frac{\gamma k_BT}{M}$, where $k_B$ is the Boltzman's constant and M is the molar mass of the gas, $\gamma$ is the adiabatic index and T is the temperature.
$v_{sound} H_2 = \large \sqrt \frac{\gamma k_BT}{M} = \large \sqrt \frac{7 k_BT}{5 \times 2.01}$
$v_{sound} CO_2 = \large \sqrt \frac{\gamma k_BT}{M} = \large \sqrt \frac{1.28 k_BT_0}{44}$
$v_{sound} H_2 = v_{sound} CO_2 \rightarrow \large \sqrt \frac{7 k_BT}{5 \times 2.01}$$= \large \sqrt \frac{1.28 k_BT_0}{44} \Rightarrow T_0 = \large\frac{7 \times 44 \times T}{5 \times 2.01 \times 1.28} \Rightarrow T_0 = \large\frac{308 \times T}{12.864}$$ \approx 24 T$
answered Jan 24, 2014

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