Question

sample gas has a pressure of 6.8 kPa at 539K. If the temperature decreases to 211K, then what will be the new pressure?

Answer 1

Gay-Lussac's Law-

`\:\:\:\:\:\:\star\longrightarrow \underline{\sf \boxed{\sf (P_1)/(T_1)=(P_2)/(T_2)}}`

`\:\:\:\:\:\:\star\longrightarrow \sf \underline{P_2=(P_1 \:T_2)/(T_1)}`

Where-

• P₁ is the initial pressure.
• T₁ is the initial temperature
• P₂ is the final pressure.
• T₂ is the final temperature

As per question, we are given -

• P₁ = 6.8 KPa
• T₁ =539 K
• T₂= 211K

Now that we are given all the required values, so we can put them into the formula and solve for P₂:-

`\:\:\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf P_2=(P_1 \:T_2)/(T_1)\\`

`\:\:\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf P_2=(6.8* 211)/(539)\\`

`\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow \sf P_2 = (1434.8)/(539)\\`

`\:\:\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf P_2 = 2.661966.........\\`

`\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow \sf\underline{ P_2 = 2.7 \:KPa}\\`

Therefore, If the temperature decreases to 211K, then the new pressure will become 2.7 KPa.