One important use for pV diagrams is in calculating work. The product p V has th

Question

One important use for pV diagrams is in calculating work. The product pV has the units of Pam3=(N/m2)?m3=N?m=J; in fact, the absolute value of the work done by the gas (or on the gas) during any process equals the area under the graph corresponding to that process on the pV diagram. If the gas increases in volume, it does positive work; if the volume decreases, the gas does negative work (or, in other words, work is being done on the gas). If the volume does not change, the work done is zero.

Part A

Calculate the work W done by the gas during process 1?2.

Express your answer in terms of p0 and V0.

If 2.17m3 of a gas initially at STP is placed under a pressure of 3.65atm , the temperature of the gas rises to 37.5?C.

What is the Volume?

3. When a diver jumps into a lake, water leaks into the gap region between the diver's skin and her wetsuit, forming a water layer about 0.5 mm thick. Assume that the total surface area of the wetsuit covering the diver is about 1.0 m2, and that the water enters the suit at 16?C and is warmed by the diver to skin temperature of 35?C. The specific heat of water is 1.00 kcal/kg?C?.

Estimate how much energy (in units of candy bars = 300 kcal) is required by this heating process.

4. An organ pipe is 158cm long.

Part A

What are the fundamental and first three audible overtones if the pipe is closed at one end?

Enter your answers numerically separated by commas.

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Part B

What are the fundamental and first three audible overtones if the pipe is open at both ends?

Enter your answers numerically separated by commas.

Explanation / Answer

1. Calculate the work W done by the gas during process 1?2

Answer: W = p x deltaV = 3po x (3Vo - Vo) = 6 poVo

2. What is the Volume?

Answer:

p1V1/T1 = p2V2/T2

or, 0.987 x 2.17 / 273.15 = 3.65 x V2 / 310.65

or, V2 = 0.987 x 2.17 x 310.65 / (273.15 x 3.65)

or, V2 = 0.667 m3

3. Volume of water layer = 0.5 x 10-3 x 1 = 0.5 m3

Mass of water layer = 0.5 x 1000 kg = 500 kg

Now, Q =mC delta T

So, Q = 500 x 1 x (35 -16) = 9500 kcal

To get the units in candybars, we divide by 300 kcal, thus : 9500/300 = 31.67 candybars

4. as v=340m/s
pipe is closed at one end
closed pipe: f=(2n-1)v/4L , (v=340m/s , L=1.58m)
n=1: f=(2*1-1)*340/(4*1.58) = 53.79[Hz] 1st.harmonic, fundamental f.
n=2: f=(2*2-1)*340/(4*1.58) = 161.39[Hz] 3rd.harmonic, 1st.overtone.
n=3: f=(2*3-1)*340/(4*1.58) = 268.98[Hz] 5th.harmonic, 2nd.overtone.
n=4: f=(2*4-1)*340/(4*1.58) = 376.58[Hz] 7th.harmonic, 3rd.overtone.

pipe is open at both ends
open pipe; f=nv/2L , (v=340m/s ,L=1.58m)
n=1: f=1*340/(2*1.58) = 107.59[Hz] 1st.harmonic, fundamental.
n=2: f=2*340/(2*1.58) = 215.19[Hz] 2nd.harmonic, 1st.overtone.
n=3: f=3*340/(2*1.58) = 322.78[Hz] 3rd.harmonic, 2nd.overtone.
n=4: f=4*340/(2*1.58) = 430.38[Hz] 4th.harmonic, 3rd.overtone.

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