Question

A sample of gas has a pressure of .87atm at 25 ˚C. What is its pressure when the temperature is increased to 50˚C?

Answer 1

Gay-Lussac's Law

It states that: "At constant volume the pressure of a gas sample is directly proportional to the absolute temperature."

This means that doubling the pressure of the gas will cause its absolute temperature to double.

Mathematically this law is expressed as:

`\boxed{\large\displaystyle\text{$\begin{gathered}\sf \bf{(P_(1))/(T_(1))=(P_(2))/(T_2) \iff \ P_1T_2=P_2T1 } \end{gathered}$} }`

We have that the data is:

P₁ = 0.87 atm

T₁ = 25 °C + 273 = 298 K

T₂ = 50 °C + 273 = 323 K

P₂ = ?

We clear for the final pressure:

`\boxed{\large\displaystyle\text{$\begin{gathered}\sf \bf{P_(2)=(P_1T_2)/(T_(1)) } \end{gathered}$}}`

`\boxed{\large\displaystyle\text{$\begin{gathered}\sf \bf{P_(2)=\frac{(0.87 \ atm)\cdot(323\\ot{k})}{298\\ot{k}} } \end{gathered}$}}`

`\boxed{\boxed{\large\displaystyle\text{$\begin{gathered}\sf \bf{P_(2)=0.94 \ atm } \end{gathered}$}}}`

If the temperature increases to 50°C, its new pressure will be 0.94 atm.