The speed of sound is givenby U sound= VRT / m = VRT/M where-cP/Cv Part A What i

Question

The speed of sound is givenby U sound= VRT / m = VRT/M where-cP/Cv Part A What is the speed of sound in Ne at 1200 K Express your answer with the appropriate units. 445 Submit My Answers Give Up Incorrect: Try Again Part B What is the speed of sound in Kr at 1200 K? Express your answer with the appropriate units. 589 Submit My Answers Give Up Incorrect; Try Again Part C What is the speed of sound in Ar at 1200 K Express your answer with the appropriate units. 589 Submit My Answers Give Up Incorrect: Try Again Part D At what temperature will the speed of sound in Xe equal the speed of sound in Ne at 1000. K? Express your answer with the appropriate units. Submit My Answers Give Up Incorrect; Try Again

Explanation / Answer

All the gases are assumed to be ideal; moreover all are monoatomic gases and hence, we will have

= Cp/Cv = (5/2*R)/(3/2*R) = 5/3 = 1.66.

Part A

Molar mass of Ne is M = 20.1797 g/mol = (20.1797 g/mol)*(1 kg/1000 g) = 0.0201797 kg/mol; R = 8.314 kg m2 s-2 K-1 mol-1 and T = 1200 K. Plug in values and obtain

vsound = *R*T/M = (1.66)*(8.314 kg m2 s-2 K-1 mol-1)*(1200 K)/( 0.0201797 kg/mol) = (820700.4068 m2 s-2) = 905.925 m/s (ans).

Part B

Molar mass of Kr is M = 83.798 g/mol = (83.798 g/mol)*(1 kg/1000 g) = 0.083798 kg/mol; R = 8.314 kg m2 s-2 K-1 mol-1 and T = 1200 K. Plug in values and obtain

vsound = *R*T/M = (1.66)*(8.314 kg m2 s-2 K-1 mol-1)*(1200 K)/( 0.083798 kg/mol) = (197635.8386 m2 s-2) = 444.562 m/s (ans).

Part C

Molar mass of Ar is M = 39.948 g/mol = (39.948 g/mol)*(1 kg/1000 g) = 0.039948 kg/mol; R = 8.314 kg m2 s-2 K-1 mol-1 and T = 1200 K. Plug in values and obtain

vsound = *R*T/M = (1.66)*(8.314 kg m2 s-2 K-1 mol-1)*(1200 K)/( 0.039948 kg/mol) = (414576.149 m2 s-2) = 643.876 m/s (ans).

Part D

First find out the speed of sound in Ne at 1000 K. We have M = 0.0201797 kg/mol and T = 1000 K; therefore,

vsound (Ne) = *R*T/M = (1.66)*(8.314 kg m2 s-2 K-1 mol-1)*(1000 K)/( 0.0201797 kg/mol) = (683917.0057 m2 s-2) = 826.993 m/s.

Next we need to find the temperature T K such that vsound(Xe) = 826.993 m/s.

Molar mass of Xe = 131.293 g/mol = (131.293 g/mol)*(1 kg/1000 g) = 0.131293 kg/mol.

We have

vsound(Xe) = *R*T/M = (1.66)*(8.314 kg m2 s-2 K-1 mol-1)*(T K)/( 0.131293 kg/mol) = (105.1178*T m2 s-2).

By the problem,

(105.1178*T m2 s-2) = 826.993 m/s

====> 105.1178*T m2 s-2 = 683917.422 m2 s-2

====> T = 6506.20

The required temperature is 6506.20 K (ans).

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