At conditions of 1.5 atm of pressure and 15.0 °C temperature, a gas occupies a volume of 45.5 mL. What will be the volume of the same gas at 1.1 atm and 30.0 °C?

At conditions of 1.5 atm of pressure and 15.0 °C temperature, a gas occupies a volume of 45.5 mL. What will be the volume of the same gas at 1.1 atm and 30.0 °C?

Feb 27, 2018

$\text{Well...what does the old combined gas equation say...?}$

Explanation:

...that $\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$...for a GIVEN molar quantity of gas....the which scenario pertains here. As usual we use $\text{absolute temperatures}$.

And so we solve for ${V}_{2} = \frac{{P}_{1} {V}_{1}}{T} _ 1 \times {T}_{2} / {P}_{2}$

$\frac{1.5 \cdot a t m \times 45.5 \cdot m L}{288.15 \cdot K} \times \frac{303.15 \cdot K}{1.1 \cdot a t m} = 65.3 \cdot m L$..

That volume has increased is consistent with the pressure drop, and temperature increase...the question should have specified that the gas was confined to a piston....with variable volume...

Feb 27, 2018

$\text{65.3 mL}$

Explanation:

You'll be using the Combined Gas Law since your question involves all three variables:

$\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$

If ($P =$ pressure , $V =$ volume, $T =$ temperature), then we know in the equation that:

• ${P}_{1} = \text{1.5 atm}$
• ${P}_{2} = \text{1.1 atm}$
• ${V}_{1} = \text{45.5 mL}$
• V_2 = ?
• ${T}_{1} = {15}^{\circ} \text{C}$
• ${T}_{2} = {30}^{\circ} \text{C}$

Before plugging in all values to solve for the volume, temperature MUST ALWAYS be in Kelvins, at least for gases. To convert Celsius to Kelvin, all you do is add $273$. So,

${T}_{1} = \text{288 K }$ and $\text{ "T_2 = "303 K}$

Now we're ready to plug and chug!

Work for calculation:

$\frac{1.5 \cdot 45.5}{288} = \frac{1.1 {V}_{2}}{303}$

$0.24 = \frac{1.1 {V}_{2}}{303}$

$71.8 = 1.1 {V}_{2}$

$65.3 = {V}_{2}$

or

${V}_{2} = 65.3$

So that means the volume of the gas would be about $\text{65.3 mL}$.