Simultaneous doubling of the absolute temperature of a gas and reduction of its pressure to half, will result in __________ in the volume of the gas.
A. No change
B. Doubling
C. $${\frac{{\text{L}}}{4}^{{\text{th}}}}$$ reduction
D. Four fold increase
Answer: Option D
Solution (By Examveda Team)
According to the ideal gas law, for a fixed amount of gas:PV = nRT
Since the amount of gas (n) and gas constant (R) stay constant, the relationship between volume, temperature, and pressure is:
V ∝ T/P
You are doubling the absolute temperature (T₂ = 2T₁), and at the same time reducing the pressure to half (P₂ = 0.5P₁).
So, the new volume compared to the original volume will be:
V₂/V₁ = (T₂/T₁) * (P₁/P₂)
Substituting the values gives:
V₂/V₁ = (2T₁/T₁) * (P₁/0.5P₁)
V₂/V₁ = 2 * 2 = 4
Therefore, the volume of the gas will experience a four fold increase, not just doubling.
This Question Belongs to Chemical Engineering >> Stoichiometry
We use the Ideal Gas Law:
𝑃
𝑉
=
𝑛
𝑅
𝑇
PV=nRT
Let’s analyze what happens when:
Temperature is doubled →
𝑇
2
=
2
𝑇
1
T
2
=2T
1
Pressure is halved →
𝑃
2
=
1
2
𝑃
1
P
2
=
2
1
P
1
We assume amount of gas (n) and gas constant (R) remain constant.
Step-by-step:
From ideal gas law:
𝑃
1
𝑉
1
𝑇
1
=
𝑃
2
𝑉
2
𝑇
2
T
1
P
1
V
1
=
T
2
P
2
V
2
Substitute the changes:
𝑃
1
𝑉
1
𝑇
1
=
(
𝑃
1
/
2
)
𝑉
2
2
𝑇
1
T
1
P
1
V
1
=
2T
1
(P
1
/2)V
2
Multiply both sides to eliminate denominators:
𝑃
1
𝑉
1
=
𝑃
1
2
⋅
𝑉
2
2
=
𝑃
1
𝑉
2
4
P
1
V
1
=
2
P
1
⋅
2
V
2
=
4
P
1
V
2
Now solve for
𝑉
2
V
2
:
𝑃
1
𝑉
1
=
𝑃
1
𝑉
2
4
⇒
𝑉
2
=
4
𝑉
1
P
1
V
1
=
4
P
1
V
2
⇒V
2
=4V
1
✅ Correct answer: D. Four fold increase