A gas at a pressure of 5 atm is heated from 0° to 546°c and is simultaneously compressed to one third of its original volume. Find the final pressure of the gas.
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Given:
P₁= 5atm
T₁ = 0°C = 273 k
Let the initial volume be, V₁
P₂ = ?
T₂ = 546°C = 819 k
V₂ = 1/3 of V₁= V₁/3
From Ideal gas equation,
P₁V₁/T₁ = P₂V₂/T₂
On Substituting:
→ 5 x V₁/273 = P₂ x P₁(V₁/3)/819
→ 5V₁/273 = P₂V₁/3 x 819
→ 5V₁/273 = P₂V₁/2457
→ P₂V₁ = 5V₁/273 x 2457
→ P₂V₁ = 12285V₁/273
Cancel the term V₁ from both sides,
→ P₂ = 12285/273
→ P₂ = 45
∴ Final pressure of the gas is 45atm.
Answered by
9
Given:
P₁= 5atm
T₁ = 0°C = 273 k
Let the initial volume be, V₁
P₂ = ?
T₂ = 546°C = 819 k
V₂ = 1/3 of V₁= V₁/3
From Ideal gas equation,
P₁V₁/T₁ = P₂V₂/T₂
On Substituting:
→ 5 x V₁/273 = P₂ x P₁(V₁/3)/819
→ 5V₁/273 = P₂V₁/3 x 819
→ 5V₁/273 = P₂V₁/2457
→ P₂V₁ = 5V₁/273 x 2457
→ P₂V₁ = 12285V₁/273
Cancel the term V₁ from both sides,
→ P₂ = 12285/273
→ P₂ = 45
∴ Final pressure of the gas is 45atm.
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