Question

A sample of helium has a volume of 3.20x10^2 mL at STP. What will be its new volume (inL) if the temperature is increased to 425.0 K and its pressure is increased to 3.50 atm?

Answer 1

Let's see that the STP represents the conditions for the temperature of 0°C (273 K) and for the pressure of 1 atm.

We have this initial data and a volume of 3.20 x 10 ^(2) mL. To solve this problem, we need to use the ideal gas formula:

`(P_1V_1)/(T_1)=(P_2V_2)/(T_2),`

where T is temperature, P is pressure, and V volume. Subindex 1 is the initial data and subindex 2 is the final data.

We want to find the final volume, so clearing for V2 in the formula, we're going to obtain:

`V_2=(P_1V_1T_2)/(T_1P_2)\text{.}`

And the final step is replacing the data that we have, where the final data is 425.0 K and 3.50 atm (remember that the volume must be in liters, 1 liter is 1000 mL, so 3.20 x 10^2 mL is 0.32 L):

`\begin{gathered} V_2=\frac{1\text{ atm }\cdot\text{ 0}.32\text{ L }\cdot425.0K}{273\text{ K }\cdot\text{ 3.50 atm}}, \\ V_2=0.14\text{ L.} \end{gathered}`

The answer is that the new volume of the sample of helium would be 0.14 L.