Answer:
Explanation:
To solve this problem, we can use the combined gas law, which states:
�
1
×
�
1
�
1
=
�
2
×
�
2
�
2
T
1
​
P
1
​
×V
1
​
​
=
T
2
​
P
2
​
×V
2
​
​
Where:
�
1
P
1
​
 and
�
2
P
2
​
 are the initial and final pressures, respectively.
�
1
V
1
​
 and
�
2
V
2
​
 are the initial and final volumes, respectively.
�
1
T
1
​
 and
�
2
T
2
​
 are the initial and final temperatures in Kelvin, respectively.
First, we need to convert the temperatures to Kelvin using the formula
�
(
�
)
=
�
(
°
�
)
+
273.15
T(K)=T(°C)+273.15:
Given:
�
1
=
2.7
 
L
V
1
​
=2.7L
�
1
=
29.4
 
psi
P
1
​
=29.4psi
�
1
=
27
°
�
=
27
+
273.15
=
300.15
 
K
T
1
​
=27°C=27+273.15=300.15K
�
2
=
58.8
 
psi
P
2
​
=58.8psi
�
2
=
12
°
�
=
12
+
273.15
=
285.15
 
K
T
2
​
=12°C=12+273.15=285.15K
Now, let's rearrange the equation to solve for
�
2
V
2
​
:
�
2
=
�
1
×
�
1
×
�
2
�
2
×
�
1
V
2
​
=
P
2
​
×T
1
​
P
1
​
×V
1
​
×T
2
​
​
Now plug in the values:
�
2
=
29.4
 
psi
×
2.7
 
L
×
285.15
 
K
58.8
 
psi
×
300.15
 
K
V
2
​
=
58.8psi×300.15K
29.4psi×2.7L×285.15K
​
�
2
≈
2291.6539
17628.492
 
L
V
2
​
≈
17628.492
2291.6539
​
L
�
2
≈
0.130
 
L
V
2
​
≈0.130L
Rounding to two significant figures, the volume of the oxygen in the tank at the new conditions is approximately
0.13
 
L
0.13L.
