Question

a sample of gas has the volume of 10L at 20degree C . for the volume to become 30L at a constant pressure what will be Celsius temperature have to be
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Answer 1

Answer:-

Given:

Initial Volume (V₁) = 10 L

Initial Temperature (T₁) = 20° C = 20 + 273 = 293° K

Final Volume (V₂) = 30 L

Let the final Temperature be T₂.

We know that,

Charle's law of gases states that at constant pressure the Volume of a gas is directly proportional to the temperature.

⟶ V ∝ T

⟶ V/T = constant

⟶ V₁/T₁ = V₂/T₂

⟶ 10/293 = 30/T₂

⟶ T₂ (10) = 30 * 293

⟶ T₂ = (30)(293)/10

⟶ T₂ = 879° K

⟶ T₂ = 879 - 273

⟶ T₂ = 606° C

∴ The temperature of the gas increases to 606° C.

Answer 2

$\huge{\boxed{\rm{\red{Question}}}}$

A sample of gas has the volume of 10L at 20degree C . for the volume to become 30L at a constant pressure what will be Celsius temperature have to be

$\huge{\boxed{\rm{\red{Answer}}}}$

${\bigstar}\large{\boxed{\sf{\pink{Given \: that}}}}$

• Initial temperature = 20°C = 20 + 273 = 293° K
• Initial volume = 10 litres.
• Final volume = 30 litres

$\large\green{\texttt{Let the final temperature be T2}}$

${\bigstar}\large{\boxed{\sf{\pink{We\: know\: that}}}}$

★ Charle's gas law states that at constant pressure the volume of gas is directly proportional to temperature.

☞ $\large\purple{\texttt{V∞T}}$

☞ $\large\purple{\texttt{V/T = Constant}}$

☞ $\large\purple{\texttt{V1/T1 = V2/T2}}$

☞ $\large\purple{\texttt{10/293 = 30/T2}}$

☞ $\large\purple{\texttt{T2(10) = 30 × 293}}$

☞ $\large\purple{\texttt{T2 = (30)(293)/10}}$

☞ $\large\purple{\texttt{T2 = 879°K}}$

☞ $\large\purple{\texttt{T2 = 879 - 273}}$

☞ $\large\purple{\texttt{T2 = 606°C}}$

The increased temperature of gas is $\large\red{\texttt{606°C}}$

@Itzbeautyqueen23