# Example Of Applied Physics I Case Study

Published: 2021-07-14 11:25:06

Category: Law, Leadership, Experiment, Stress, Gas, Ideal, Ideal Gas, Ideal Gas Law

Type of paper: Essay

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Heat and Thermodynamics
Experiment 1:
Isobaric (constant pressure) Process
P1 = P2 = 100 kPa

The blanks have been filled in red.
in
by
out
on
in
in
According to ideal gas law, PV = nRT
For a constant mass, isobaric process, V/T = nR/P = constant.
Hence, V1/T1 = V2/T2.
For each of the cases, V1/T1 = 1/300 dm3/K. V2/T2 has been calculated in the table below for each case.
3.33 x 10-3
3.33 x 10-3
3.33 x 10-3
3.33 x 10-3
3.33 x 10-3
3.33 x 10-3
3.33 x 10-3
3.33 x 10-3
3.33 x 10-3
3.33 x 10-3
3.33 x 10-3
3.33 x 10-3
Hence, within the limits of rounding errors, the ideal gas law is verified for the isobaric case.
Experiment 2:
Isochoric (constant volume) process
V1 = V2 = 1.00 dm3

The blanks have been filled in red.
in
in
out
in
in
According to ideal gas law, PV = nRT
For a constant mass, isochoric process, P/T = nR/V = constant.
Hence, P1/T1 = P2/T2.
For each of the cases, P1/T1 = 1/3 kPa/K. P2/T2 has been calculated in the table below for each case.
Hence, within the limits of rounding errors, the ideal gas law is verified for the isochoric case.
Discussion: For a constant volume process, work done is 0 since the volume does not change. Increase in temperature leads to an increased pressure, which is established only through supply of external heat. Similarly, removing heat leads to a reduced temperature and pressure, but the work done is 0.
Experiment 3:
Isothermal (constant temperature) process
T1 = T2 = 300 K
The blanks have been filled in red.
According to ideal gas law, PV = nRT
For a constant mass, isothermal process, PV = nRT = constant
Hence, P1V1 = P2V2
For each of the cases, P1V1 = 100 kPa.dm3. P2V2 has been calculated in the table below for each case.
Hence, within the limits of rounding errors, the ideal gas law is verified for the isothermal case.

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