Question

A gas has a pressure of 6.58 kPa at 540 K. What will the pressure be at 210 K if the volume does not change?

Answer 1

$\huge{\underline{\pink{\tt{Given,}}}}$

• A gas has a Pressure (Initial Pressure) = 6.58kPa
• A Gas has Pressure of 6.58kPa at (Initial Temperature) = 540K
• Final Temperature = 210K

$\huge{\underline{\pink{\tt{To \ Find,}}}}$

• If the volume does not change then The Pressure will be at 210K

$\huge{\underline{\pink{\tt{Solution :}}}}$

$\longrightarrow$ We know that Gay Lussac's Law :

$\bigstar \boxed{\sf{\red{\dfrac{P_{1}}{T_{1}} = \dfrac{P_{2}}{T_{2}}}}}$

Here,

• $\sf{P_{1}}$ - Initial Pressure
• $\sf{P_{2}}$ - Final Pressure
• $\sf{T_{1}}$ - Initial Temperature
• $\sf{T_{2}}$ - Final Temperature

$\huge{\underline{\purple{\mathfrak{Gay\:Lussac's\:Law:}}}}$

$\longrightarrow$ Gay - Lussac's Law Defined that the Pressure of the Fixed mass of gas  is Directly Proportional to the Temperature of the Gas at the Constant Volume .

$\bigstar \boxed{\sf{P \propto T}}$

$\underline{\underline{\pink{\tt{Now,According\:to\:the\:Question :}}}}$

$\implies \sf{\dfrac{P_{1}}{T_{1}} = \dfrac{P_{2}}{T_{2}}}$

$\implies \sf{P_{2} = \dfrac{T_{2}}{T_{1}} \times P_{1}}$

$\implies \sf{P_{2} = \dfrac{210}{540} \times 6.58}$

$\implies \sf{P_{2} = \dfrac{21}{54} \times \dfrac{658}{100}}$

$\implies \sf{P_{2} = \dfrac{13818}{5400}}$

$\implies \boxed{\sf{P_{2} = 2.558kPa}}$ (Answer)

$\rule{200}2$