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PHYSICS
Case A
Figure 14
1
st
phase
: assuming that the pressure remains constant. Using Gay-Lussac’s first law calculate the volume
V
2
of
the gaswhich is heated from a temperature
t
1
=20+273=293
K
to a temperature
t
2
=50+273=323
K
.
V
1
:
V
2
=
t
1
:
t
2
V
1
*
t
2
=
V
2
*
t
1
V
2
=
V
1
* t
2
t
1
V
2
=
0,98
[dm
3
]
* 323
[K]
V
2
=
1,08
dm
3
293
[K]
Due to only the increase of the temperature of the gas, the volume has increased
V
2
=
1,08
dm
3
.
2
nd
phase
: we observe the behaviour of the volume
V
2
with an increase of double the Force applied on the piston.
F
2
= 2 * F
1
F
2
= 2 * 980
N
F
2
=
1960
N
As a result of the increase in the Force
F
2
, applied to the piston, there is a reduction in the volume
V
3
.
V
2
:
V
3
=
F
2
:
F
1
V
3
=
V
2
* F
1
F
2
V
3
=
1,08
[dm
3
]
* 980
[N]
V
3
=
0,54
dm
3
1960
[N]
Under the action of the Force
F
2
and temperature
t
2
the volume is reduceduntil it reaches
V
3
=
0,54
dm
3
.
Case B
The unit of measurement of the pressure is
Kg/cm
2
, it is possible to calculate the value of the pressure using the
value of the load
F
1
,
F
2
and the surface of the piston.
Figure 14
1
st
phase
: calculation of the area of the piston
S
=
r * r *
π
S
=
25
*
25
*
3,14
S
=
1962,5
mm
2
S
=
19,6
cm
2
Calculate the initial pressure
p
1
:
p
1
=
F
1
980
[N]
≅
50
N
⁄
cm
2
p
1
≅
5Kg
⁄
cm
2
S
19.6
[cm
2
]
Figure 14
2
nd
phase
: calculation of the final pressure
p
2
:
p
2
=
F
2
1960
[N]
≅
100
N
⁄
cm
2
p
2
≅
10Kg
⁄
cm
2
S
19.6
[cm
2
]
Load and pressure are directly proportional, using Boyle’s law and substituting the known values gives:
V
2
:
V
3
=
p
2
:
p
1
V
2
*
p
1
=
p
2
*
V
3
V
3
=
V
2
* p
1
p
2
V
3
=
1,08
[dm
3
]
* 50
[N / cm
2
]
V
3
=
0,54
dm
3
100
[N / cm
2
]
In both cases the result is
V
3
=
0,54
dm
3
.
1
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