A gas balloon has volume of 106 litres when the temperature is 42 deg celsius and pressure is 740.0 mm of mercury. What will its volume be at 20 deg celsius and 780 mm of mercury pressure?
Answers
Answered by
0
Answer:
Applying ideal gas equation considering the gas inside the balloon to be ideal,to get the new volume,
So,from ideal gas law,we get,
P
V
T
= constant
so,we can say,
P
V
T
=
P
'
V
'
T
'
Where,
P
=
740
m
m
o
f
H
g
,
P
'
=
780
m
m
o
f
H
g
,
T
=
(
45
+
273
)
=
318
K
,
T
'
=
(
20
+
273
)
=
293
K
,
V
=
106
L
So,
V
'
=
92.66
L
Answered by
139
Answer:
Required volume is 93.6 Litres.
Explanation :
We have :
→ Initial volume (V₁) = 106 L
→ Initial temperature (T₁) = 42°C
→ Initial pressure (P₁) = 740 mm Hg
→ Final temperature (T₂) = 20°C
→ Final pressure (P₂) = 780 mm Hg
Firstly, let's convert the units of initial and
final temperature from °C to K.
Initial temperature:
→ 0°C = 273 K
→ 42°C = 273 +42
⇒ 315 K
Final temperature :
→ 20°C = 273 + 20
→ 293 K
Now, putting the values in the "Combined
Gas Equation", we get :
P₁V₁/T₁= P₂V₂/T₂
(740 × 106)/315 = (780 × V₂)/293
→ 78440/315 = 2.66V₂
249 = 2.66V2
V₂ = 249/2.66
⇒ V₂ = 93.6 L
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