Question

1. A long cylindrical gas-tight tube, with a sound source at one end and sensor at the other, was constructed to measure the speed by recording the time it take a sound wave to travel the length of the tube (see diagram). 

i. Initially, the tube was filled with air at atmospheric pressure (P=1.012x105Pa & T=295.2K). When a sound wave is produced by the source, the elapsed time for the wave to reach the sensor was 3.64x10-3s. Determine the length of the gas tube (i.e. distance between source and sensor). Hint: You will first need to calculate the speed of sound at this temperature. 

ii. The air was then removed from the tube and it was filled with a diatomic gas. In a series of experiments, the speed of sound was measured in the gas at various gas temperatures. The wave speed vs temperature graph for the experiment is shown below, as well as the corresponding curve fit generated. 

a. Express value of the fit constant "A" in the graph in SI units. Include the units in v  your answer.

b. Use the fit constant "A" to estimate the average mass of one of the gas particles inside the container. Note: Use the accepted value for Boltzmann Constant. 

Answer 1

ANSWER :
ره (1) Griven that P21.012 Xiospa T= 295.26 t. 3.64x10-35 Uring Laplace" correction formula وا 8P op الیا knowl Sau 7 1-293kg m-3. 8:1.41 Cfur au] Standard valu. Putin مباحا) ger 1 1:41 X 1.012 X10 لا 332.5ms I.293 Now to delermini Length وا L t L=uxta 332.5 X 3.64 X10-3 1.2103 m AM was then removed and filled with dialomic gas, وسيا know یا &P using gas equahon PU=RT

know densing a Mars volume Volume (U12 2/2 SO PM RT la cia RT M Uz re P so 8 RT unit less gass Constant. Unit is isky SO ham Comparining U SAT JR 393.7 3.783 Units y Ra kg m² k'' molts-2 Kg A= kg/m² qc mol's-2 m² kl mol's-2 iça

(61 Now determine mass of one molecule of gas ERT Mam NA m=ausmass of I molecule NAZ Avog adrw No. we know R = NAKB KB 21-38x10-23 JK". : وا ERT So Az &R A 8 kB ER MNA -23 8 kB 1.4 x 1138x10-? -26 4.907x10 'kg 393.7 average mass of one of the gas particles.

Answer 2

I have solved your problem which is shown below:

ره (1) Griven that P21.012 Xiospa T= 295.26 t. 3.64x10-35 Uring Laplace" correction formula وا 8P op الیا knowl Sau 7 1-293kg m-3. 8:1.41 Cfur au] Standard valu. Putin مباحا) ger 1 1:41 X 1.012 X10 لا 332.5ms I.293 Now to delermini Length وا L t L=uxta 332.5 X 3.64 X10-3 1.2103 m AM was then removed and filled with dialomic gas, وسيا know یا &P using gas equahon PU=RT

know densing a Mars volume Volume (U12 2/2 SO PM RT la cia RT M Uz re P so 8 RT unit less gass Constant. Unit is isky SO ham Comparining U SAT JR 393.7 3.783 Units y Ra kg m² k'' molts-2 Kg A= kg/m² qc mol's-2 m² kl mol's-2 iça

(61 Now determine mass of one molecule of gas ERT Mam NA m=ausmass of I molecule NAZ Avog adrw No. we know R = NAKB KB 21-38x10-23 JK". : وا ERT So Az &R A 8 kB ER MNA -23 8 kB 1.4 x 1138x10-? -26 4.907x10 'kg 393.7 average mass of one of the gas particles.

In case you have any query please write in a comment. I hope you like it. Thank you