A certain mass of gas occupies a volume of 2 litres at pressure of 1 atmospheres and a temperature of 27°C. Keeping pressure constant the volume would be doubled by raising the temperature to?
Answers
Answer :-
Required temperature is 327°C .
Explanation :-
We have :-
→ Initial volume (V₁) = 2 L
→ Initial temperature (T₁) = 27°C
→ Final volume (V₂) = 2(2) = 4 L
→ Pressure remains constant (1 atm) .
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Firstly, let's convert the initial temperature from °C to K .
⇒ 0°C = 273 K
⇒ 27°C = 273 + 27
⇒ 300 K
Now according to Charle's Law, the required temperature will be given by :-
V₁/T₁ = V₂/T₂
⇒ 2/300 = 4/T₂
⇒ 2T₂ = 4(300)
⇒ T₂ = 1200/2
⇒ T₂ = 600 K
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Required temperature in °C is :-
⇒ 273 K = 0°C
⇒ 600 K = (600 - 273)°C
⇒ 327°C
REQUIRED EXPLANATION :-
- Initial volume is 2 litres
- By keeping pressure constant Volume is doubled i.e
Final volume will be 2(2) l = 4 litres
- Initial temperature is 27°C
i.e
- v₁ = 2 l
- v₂ = 4 l
- t₁ = 27°C
- t₂ = ?
Pressure is constant
To find :-
- Final Temperature .
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So,
We can find the final Temperature by using gas law called Charle's law that is :-
Charles law :-
At constant pressure P as volume increases , temperature also increases.
V₁ / T₁ = V ₂/T₂
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Temperature must be in kelvin for all gas laws. So we are used to convert the Celsius- Kelvins
As they given in Celsius we have to convert into kelvins.
C= K-273
=> 27 = K-273
=> 27 + 273 = K
K = 300
So, the initial temperature is 300K
Substituting the values,
=> V₁ / T₁ = V ₂/T₂
=> 2/300 = 4/T ₂
=> 1/300 = 2/T₂
=> T₂ = 300×2 K
T₂ = 600K
So, the Raising Temperature is 600K
Temperature in Celsius :-
We already know the relation between Kelvins and Celsius
C = K -273
=> C = 600-273
=>C = 327
So, the final temperature(T₂) is 327°C
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Know more about Gas laws . . .
P₁ V₁ = P₂ V ₂ ~ Boyle's law
PV = nRT ~Ideal gas law
P₁ V₁/T₁ = P₂ V ₂/T₂ ~Combined law
P₁ /T₁ = P₂ /T ₂ ~ Gay-Lussacs law