Simultaneous doubling of the absolute temperature of a gas and reduction...

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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

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Comments (1)

Atanu Chatterjee:

6 months ago

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