Question

If 12L of neon at 23 °C is allowed to expand to 52 L, what must the new temperature be to maintain constant 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=12 L
• T₁ = 23°C
• V₂ =52 L

We are given the initial in °C.So, we first have to convert the temperature in Celsius to kelvin by adding 273-

`\:\:\:\:\:\:\star\sf T_1` = 23+ 273 =296K

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=(296)/(12) * 52\\`

`\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf T_2=24.66....... * 52\\`

`\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf T_2=1282.66.........\:K\\`

`\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf T_2=(1282.67 -273)°C\\`

`\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf T_2=1009.66....\:°C\\`

`\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf \underline{T_2=1009.67\:°C}\\`

Therefore, If 12L of neon at 23 °C is allowed to expand to 52 L then the new temperature will become 1009.67°C or 1282.67K to maintain constant pressure.