Question

A gas occupies volume of 0.35 $d{m}^3$ at 290 K and 92.458K $N{m}^{-2}$pressure. Calculate the volume of gas at STP (Ans : 0.3 $d{m}^3$)

Answer 1

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

.Pressure- 100 Kpa

.Temperature-$0^c$

since no mention about 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 are not familiar with the combined gas law. you can derive it derive it using the ideal gas law.

Notice that you will need to convert from Kpa to $N{m}^{-2}$ by using conversion factor

$1pa=1N{m}^{-2}$

so lets say that for the initial state of gas, you can write

$P_2\times V_2$=$n\times R\times T_2$

if you divide these two equations, you will get the combined gas law from

$\frac{P_1{V}_1}{P_2{V}_2}$=$\frac{n\times R\times T_1}{n\times R\times T_2}$ $\iff \frac{P_1{V}_1}{T_1}$=$\frac{P_2{V}_2}{T_2}$

Rearrange to solve for $V_2$

$V_2$=$\frac{P_1}{P_2}\times \frac{T_2}{T_1}\times V_1$

Plug in your values- do not forget about converting the temperature from degrees

$V_2$=$\frac{92.558\times 10N{m}^{-2}}{100\times {10}^{3}N{m}^{-2}}\times \frac{(273.15+0)K}{290K}\times 0.35d{m}^3$

$V_2=0.3051d{m}^3$

$v_2=0.31d{m}^3$