For the first one, W = 7.3 kJ, Q = 25.83kJ
For the second one, W = Q = 124KJ
First One:
Parameters given are:
Mass m = 0.05kg
[tex]V_{2}[/tex] = 0.0658 [tex]m^{3}[/tex]
Since the pressure is constant, [tex]P_{1} = P_{2} = 2bar[/tex]
While [tex]T_{1} = 130[/tex]°C
Convert degree Celsius to degree kelvin
Temperature = 273 + 130 = 403 K
Before we can get the work done, we need to calculate the initial volume [tex]V_{1}[/tex] by using Ideal gas general formula.
[tex]P_{1}V_{1} = mRT_{1}[/tex]
Make [tex]V_{1}[/tex] the subject of formula
[tex]V_{1}[/tex] = [tex]mRT_{1}[/tex] / [tex]P_{1}[/tex]
[tex]V_{1}[/tex] = (0.05 x 287 x 403) / (2 x [tex]10^{5}[/tex])
[tex]V_{1}[/tex] = 0.0289[tex]m^{3}[/tex]
From the table, specific volumes are:
[tex]u_{1}[/tex] = 0.578[tex]m^{3}[/tex]kg
[tex]u_{2}[/tex] = 1.316[tex]m^{3}[/tex]kg
Work done can be calculated by using the below formula
W = P( [tex]u_{2}[/tex] - [tex]u_{1}[/tex] )
W = 2 x [tex]10^{5}[/tex]( 1.316 - 0.578)
W = 147600 J/kg
The Total work done = 147600 x 0.05 = 7380J = 7.3 kJ
To calculate the heat supplied by using the heat formula
Q = m[tex]C_{p}[/tex]( [tex]T_{2} - T_{1}[/tex]), we need to calculate for final temperature [tex]T_{2}[/tex] and also check the table for
where [tex]C_{p}[/tex] = 1.005
we can calculate for [tex]T_{2}[/tex] by using Ideal gas formula
[tex]P_{2}V_{2} = mRT_{2}[/tex]
make [tex]T_{2}[/tex] the subject of formula
[tex]T_{2}[/tex] = PV/mR
[tex]T_{2}[/tex] = (2 x [tex]10^{5}[/tex] x 0.0658) / (0.05 x 287)
[tex]T_{2}[/tex] = 917 K
Substituting [tex]T_{2}[/tex] and other parameters into the formula
Q = m[tex]C_{p}[/tex]([tex]T_{2} - T_{1}[/tex])
Q = 0.05 x 1.005 ( 917 - 403)
Q = 25.83 KJ
Second one
From the question, the following parameters are given
mass m = 1kg
Molar M = 28kg/mol
Since the system is compressed reversibly and isothermally,
[tex]T_{2}[/tex] = T = 20°C
[tex]P_{1}[/tex] = 1.01 bar
[tex]P_{2}[/tex] = 4.2 bar
To calculate both heat and work done, we will use the formula below
W = RTln[tex]P_{1}[/tex]/[tex]P_{2}[/tex]
But R = Ro/M
R = 8314/28
R = 297 J/kg.k
W = 297 x 293 x ln(1.01/4.2)
W = - 124kJ / kg
We can therefore conclude that the work input during the process is
124 KJ While the heat produced is also 124 KJ because W = Q
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