Oxygen gas is commonly sold in 48.0L steel containers at a pressure of 130atm .What volume (in liters) would the gas occupy if its temperature was raised from 20.5∘C to 40.0∘C at constant 130atm ?

If it's still in that 48.0 L container the volume will be, guess what, 48.0L. I assume you are asking what volume if the gas were allowed to expand freely but the question isn't clear on that.

(V1/T1) = (V2/T2)
Remember T must be in kelvin.

To solve this problem, you can use the ideal gas law, which states that PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T is temperature in Kelvin.

First, let's convert the temperatures from Celsius to Kelvin. The formula to convert Celsius to Kelvin is T(K) = T(°C) + 273.15.

Initial temperature, T1 = 20.5 °C + 273.15 = 293.65 K
Final temperature, T2 = 40.0 °C + 273.15 = 313.15 K

Since the pressure remains constant, we can simplify the ideal gas law equation to V1/T1 = V2/T2.

Now, we can plug in the given values:

V1 = 48.0 L (initial volume)
T1 = 293.65 K (initial temperature)
T2 = 313.15 K (final temperature)

V2 is the unknown volume.

Now, we can rearrange the equation to solve for V2:

V2 = (V1 * T2) / T1

Substituting the known values:

V2 = (48.0 L * 313.15 K) / 293.65 K
V2 = 51.14 L

Therefore, the volume of the gas would be approximately 51.14 liters if its temperature was raised from 20.5 °C to 40.0 °C at a constant pressure of 130 atm.