Question

When the volume of a gas is

changed from 3.75 L to 6.52 L,
the temperature will change from
65.0 °C to
°C.
T = [?] °C
Assume that the number of moles and the pressure
remain constant. Be careful of the temperature units.
Temperature (8CY

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=3. 75 L
• T₁ = 65°C
• V₂ =6.52 L

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

`\:\:\:\:\:\:\star\sf T_1` = 65+ 273 =338 K

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=(338)/(3.75) * 6.52\\`

`\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf T_2=90.13333...... * 6.52\\`

`\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf T_2=587.669.........\:K\\`

`\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf T_2=(587.67 -273)°C\\`

`\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf T_2=314.66933…....\:°C\\`

`\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf \underline{T_2=314.67\:°C}\\`

Therefore, the temperature will change from 65°C to 314.67°C, when the volume of a gas is changed from 3.75 L to 6.52 L.