Question

What change in volume results if 40 mL of gas is cooled from 33 °C to 5 °C?

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=40 mL
• T₁ = 33°C
• T₂ =5°C

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

`\:\:\:\:\:\:\star\sf T_1` = 33+ 273 = 306K

`\:\:\:\:\:\:\star\sf T_2` =5+273 = 278K

Now that we have obtained all the required values, so we can put them into the formula and solve for V₂ :-

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

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

`\:\:\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf V_2= (40)/(306)* 278\\`

`\:\:\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf V_2= 0.13071...........* 278\\`

`\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow \sf V_2 = 36.33892...........\\`

`\:\:\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf\underline{ V_2 = 36.34 \:mL}\\`

Therefore, the volume will become 36.34 mL if 40 mL of gas is cooled from 33 °C to 5 °C.