If 10.0 liters of oxygen at stp are heated to 512 °c, what will be the new volume of gas if the pressure is also increased to 1520.0 mm of mercury?

If 10.0 liters of oxygen at STP are heated to 512 °C and its pressure increased to 1520.0 mmHg, the new volume will be 14.4 L.

10.0 L (V₁) of oxygen is at standard temperature (T₁ = 273.15 K) and standard pressure (P₁ = 760.0 mmHg).

It is heated to 512 °C (T₂) and the pressure increased to 1520.0 mmHg (P₂). We will convert 512 °C to Kelvin using the following expression.

Then, we can calculate the new volume (V₂) using the combined gas law.

If 10.0 liters of oxygen at STP are heated to 512 °C and its pressure increased to 1520.0 mmHg, the new volume will be 14.4 L.

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14.4 L

Explanation:

Initially, the gas is at stp (standard conditions), which means

and its initial volume is

Later, the gas is heated to a final temperature of

and the pressure is increased to

So we can use the ideal gas equation to find the new volume, V2: