Question

A steel container with a volume of 30L is filled with oxygen to a pressure of 9.00 atm at 28.0°C. What is the pressure of the temperature changes to 129.0°C

Answer 1

`\large \boxed{\text{12.0 atm}}`

Step-by-step explanation:

The volume and amount of gas are constant, so we can use Gay-Lussac’s Law:

At constant volume, the pressure exerted by a gas is directly proportional to its temperature.

\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}}

Data:

p₁ = 9.00 atm; T₁ = 28.0 °C

p₂ = ?; T₂ = 129.0 °C

Calculations:

1. Convert the temperatures to kelvins

T₁ = (28.0 + 273.15) K = 301.15

T₂ = (129.0 + 273.15) K = 402.15

2. Calculate the new pressure

`\begin{array}{rcl}(9.00)/(301.15) & = & (p_(2))/(402.15)\\\\0.02989 & = & (p_(2))/(402.15)\\\\0.02989 * 402.15 &=&p_(2)\\p_(2) & = & \textbf{12.0 atm}\end{array}\\\text{The new pressure is $\large \boxed{\textbf{12.0 atm}}$}`