Question

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.​

Answer 1

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.

Answer 2

gívєn:-

• $p 1 = 5 \: atm$
• $t1 = 0 + 273 = 273k$
• $t2 = 546 + 273 = 819k$
• $v1 = v$
• $v2 = \frac{1}{3} v$

$p2 = to \: find.$

$according \: to \: ideal \: gas \: equation, \:$

$\boxed{\frac{p1 \times v1}{t1} = \frac{p2 \times v2}{t2} }$

$p2 = \frac{p1 \times v1 \times t2}{t1 \times v2}$

$p2 = \frac{5 \times v \times 819}{273 \times \frac{v}{3} }$

$p2 = \frac{5 \times v \times 819 \times 3}{273 \times v}$

$p2 = 45 \: atm$