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.