A gas occupies volume of 0.35dm3 at 290K and 92.558K Nm-2 pressure. Calculate the volume of gas at STP?

Oct 18, 2015

${\text{0.31 dm}}^{3}$

Explanation:

The first thing to do is write out what the STP conditions are

• Pressure - $\text{100 kPa}$
• Temperature - ${0}^{\circ} \text{C}$

SInce no mention about the number of moles of gas was made, you can assume that it remains constant. This means that you can use the combined gas law to find the volume occupied by the gas at STP.

If you're not familiar with the combined gas law, you can derive it using the ideal gas law.

Notice that you will need to convert the pressure from $\text{kPa}$ to ${\text{N m}}^{- 2}$ by using the conversion factor

${\text{1 Pa" = "1 N m}}^{- 2}$

So, let's say that for the initial state of the gas, you can write

${P}_{1} \cdot {V}_{1} = n \cdot R \cdot {T}_{1}$

For the second state of the gas, i.e. at STP, you can write

${P}_{2} \cdot {V}_{2} = n \cdot R \cdot {T}_{2}$

If you divide these two equations, you will get the combined gas law form

$\frac{{P}_{1} {V}_{1}}{{P}_{2} {V}_{2}} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{n \cdot R}}} \cdot {T}_{1}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{n \cdot R}}} \cdot {T}_{2}} \iff \frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$

Rearrange to solve for ${V}_{2}$

${V}_{2} = {P}_{1} / {P}_{2} \cdot {T}_{2} / {T}_{1} \cdot {V}_{1}$

Plug in your values - don't forget about converting the temperature from degrees Celsius to Kelvin!

V_2 = (92.558 * 10^3color(red)(cancel(color(black)("Nm"^(-2)))))/(100 * 10^3color(red)(cancel(color(black)("Nm"^(-2))))) * ((273.15 + 0)color(red)(cancel(color(black)("K"))))/(290color(red)(cancel(color(black)("K")))) * "0.35 dm"^3

${V}_{2} = {\text{0.3051 dm}}^{3}$

Rounded to two sig figs, the number of sig figs you gave for the volume and temperatue of the gas, the answer will be

${V}_{2} = \textcolor{g r e e n}{{\text{0.31 dm}}^{3}}$