Chapter 10: Problem 1459

If the mass of 1 mole of air is $29 \times 10^{-3} \mathrm{~kg}$, then the speed of sound in it at STP is $(\gamma=7 / 5) .\left\\{\mathrm{T}=273 \mathrm{~K}, \mathrm{P}=1.01 \times 10^{5} \mathrm{~Pa}\right\\}$ (A) $270 \mathrm{~m} / \mathrm{s}$ (B) $290 \mathrm{~m} / \mathrm{s}$ (C) $330 \mathrm{~m} / \mathrm{s}$ (D) $350 \mathrm{~m} / \mathrm{s}$

Short Answer

Expert verified

The speed of sound in air at STP is approximately $330 \mathrm{~m} / \mathrm{s}$. The correct answer is (C).

Step by step solution

Since the mass of 1 mole of air is given as $29 \times 10^{-3}\mathrm{kg}$, we can calculate the gas constant $R$ using the given pressure and temperature. The formula for the gas constant is: $ PV=nRT$ For 1 mole of gas, n=1, so $ R = \frac{PV}{T}$ #Step 2: Plug in the values and calculate R#

We are given the values of P and T. So, let's plug them into the formula and solve for R: $ R = \frac{1.01\times10^5Pa}{(1)(273K)}$ Calculating R gives us: $R \approx 370.3 \mathrm{J} / \mathrm{kg}\cdot\mathrm{K}$ #Step 3: Calculate the speed of sound using the formula#

Now that we have the value of R, we can use the formula for the speed of sound in an ideal gas: $ v = \sqrt{\frac{\gamma RT}{M}}$ We know the values of γ, R, T, and M. So we can plug them into the formula: $ v = \sqrt{\frac{(7/5)(370.3)(273)}{29\times10^{-3}}}$ #Step 4: Solve for the speed of sound v#

Now we can solve for v: $v \approx 331 \mathrm{~m} / \mathrm{s}$ Comparing this to the available options, the speed of sound in air at STP is closest to 330 m/s. The correct answer is (C) $330 \mathrm{~m} / \mathrm{s}$.