If the pressure on an enclosed gas is doubled and its temperature rises from 27ºC to 127ºC, then its volume will
increase by \(\frac{2}{3^{rd}}\)
increase by \(\frac{1}{3^{rd}}\)
decrease by \(\frac{2}{3^{rd}}\)
decrease by \(\frac{1}{3^{rd}}\)
Explanation
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Discussions (13)

shouldn't it be that it decreased by 1/3rd, since the new volume is 2/3 of the old volume.

Initial temperature:
π
1
=
27
β
πΆ
=
27
+
273
=
300
β
K
T
1
β
=27
β
C=27+273=300K
Final temperature:
π
2
=
127
β
πΆ
=
127
+
273
=
400
β
K
T
2
β
=127
β
C=127+273=400K
Pressure doubles:
π
2
=
2
π
1
P
2
β
=2P
1
β
Initial volume:
π
1
V
1
β
Find: final volume
π
2
V
2
β
Step 1: Use Combined Gas Law
The Combined Gas Law (from Boyleβs & Charlesβs laws) is:
π
1
π
1
π
1
=
π
2
π
2
π
2
T
1
β
P
1
β
V
1
β
β
=
T
2
β
P
2
β
V
2
β
β
Where all temperatures are in Kelvin.
Step 2: Solve for
π
2
V
2
β
π
2
=
π
1
π
1
π
2
π
2
π
1
V
2
β
=
P
2
β
T
1
β
P
1
β
V
1
β
T
2
β
β
Substitute
π
2
=
2
π
1
P
2
β
=2P
1
β
:
π
2
=
π
1
π
1
π
2
2
π
1
π
1
=
π
1
π
2
2
π
1
V
2
β
=
2P
1
β
T
1
β
P
1
β
V
1
β
T
2
β
β
=
2T
1
β
V
1
β
T
2
β
β
π
2
=
π
1
Γ
400
2
Γ
300
=
400
600
π
1
=
2
3
π
1
V
2
β
=
2Γ300
V
1
β
Γ400
β
=
600
400
β
V
1
β
=
3
2
β
V
1
β
Answer:
π
2
=
2
3
π
1
V
2
β
=
3
2
β
V
1


