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The temperature of a gas in a closed container is 27° C. If the temperature of the gas is incresed to 300° C, then the pressure exerted is
Answer & Solution
Correct Answer:
Option
A
According to Gay-Lussac's Law, for a gas at constant volume, the pressure is directly proportional to the absolute temperature: P ∝ T.
First, convert the given temperatures to Kelvin:
Initial temperature (T₁): 27°C + 273 = 300 K
Final temperature (T₂): 300°C + 273 = 573 K
Using the formula P₂ = P₁ × (T₂ / T₁):
P₂ = P₁ × (573 / 300) ≈ P₁ × 1.91
This shows that the final pressure is approximately 1.91 times the initial pressure, which is nearly double.
Therefore, the pressure exerted is best described as doubled.
First, convert the given temperatures to Kelvin:
Initial temperature (T₁): 27°C + 273 = 300 K
Final temperature (T₂): 300°C + 273 = 573 K
Using the formula P₂ = P₁ × (T₂ / T₁):
P₂ = P₁ × (573 / 300) ≈ P₁ × 1.91
This shows that the final pressure is approximately 1.91 times the initial pressure, which is nearly double.
Therefore, the pressure exerted is best described as doubled.
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Login𝑃
1
𝑇
1
=
𝑃
2
𝑇
2
T
1
P
1
=
T
2
P
2
Where:
𝑃
1
P
1
,
𝑃
2
P
2
= initial and final pressures
𝑇
1
T
1
,
𝑇
2
T
2
= initial and final temperatures in Kelvin
Step 1: Convert Celsius to Kelvin
Initial temperature
𝑇
1
=
27
∘
𝐶
=
27
+
273
=
300
𝐾
T
1
=27
∘
C=27+273=300K
Final temperature
𝑇
2
=
300
∘
𝐶
=
300
+
273
=
573
𝐾
T
2
=300
∘
C=300+273=573K
Step 2: Use the formula
𝑃
2
𝑃
1
=
𝑇
2
𝑇
1
=
573
300
≈
1.91
P
1
P
2
=
T
1
T
2
=
300
573
≈1.91
Conclusion:
The pressure increases by approximately 1.91 times, which is nearly double.
✅ Correct answer: A.