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
A. Doubled
B. Halved
C. Trebled
D. Unpredictable
Answer: Option A
Solution (By Examveda Team)
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.
This Question Belongs to Chemical Engineering >> Stoichiometry
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Comments (1)
Related Questions on Stoichiometry
A. Potential energy
B. Intermolecular forces
C. Kinetic energy
D. Total energy
A. ΔH = 1 (+ ve)and Δ V = -ve
B. ΔH = 0
C. ΔV = 0
D. Both B and C
To solve this, we apply Gay-Lussac's Law (or the pressure-temperature relation for gases), which is:
𝑃
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.