Question

A container containing 5.75L of a gas is collected at 115 K and then allowed to expand to 25 L. What must the new temperature be to maintain the same pressure?

Answer 1

Charles's Law-

`\:\:\:\:\:\: \:\:\:\:\:\:\star\longrightarrow\sf \underline{(V_1)/(T_1)=(V_2)/(T_2)}\\`

Where:-

• V₁ = Initial volume
• T₁ = Initial temperature
• V₂ = Final volume
• T₂ = Final temperature

As per question, we are given that -

• V₁=5.75L
• T₁ = 115K
• V₂ =25 L

Now that we have obtained all the required values, so we can put them into the formula and solve for T₂ :-

`\:\:\:\:\:\: \:\:\:\:\:\:\star\longrightarrow\sf \underline{(V_1)/(T_1)=(V_2)/(T_2)}\\`

`\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf \underline{(T_2)/(V_2)=(T_1)/(V_1)}\\`

`\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf \underline{T_2=(T_1)/(V_1) * V_2}\\`

`\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf T_2=\cancel{(115)/(5.75) }* 25\\`

`\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf T_2=20 * 25\\`

`\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf T_2=500\:K\\`

`\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf T_2=(500 -273)°C\\`

`\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf \underline{T_2=227\:°C}\\`

Therefore, the new temperature will become 500K or, 227°C to maintain the same pressure.