Answer:
Approximately [tex]0.08\; \rm mol[/tex], assuming that this gas is an ideal gas.
Explanation:
Look up the standard room temperature and pressure:[tex]25\; \rm ^{\circ}C[/tex] and [tex]P = 101.325 \; \rm kPa[/tex].
The question states that the volume of this gas is [tex]V = 2\; \rm dm^{3}[/tex].
Convert the unit of all three measures to standard units:
[tex]\begin{aligned} T &= 25\; \rm ^{\circ}C \\ &= (25 + 273.15)\; \rm K \\ &= 293.15\; \rm K\end{aligned}[/tex].
[tex]\begin{aligned}P &= 101.325\; \rm kPa \\ &= 101.325 \; \rm kPa \times \frac{10^{3}\; \rm Pa}{1\; \rm kPa} \\ &= 1.01325 \times 10^{5}\; \rm Pa\end{aligned}[/tex].
[tex]\begin{aligned}V &= 2\; \rm dm^{3} \\ &= 2 \; \rm dm^{3} \times \frac{1\; \rm m^{3}}{10^{3}\; \rm dm^{3}} \\ &= 2 \times 10^{-3}\; \rm m^{3}\end{aligned}[/tex].
Look up the ideal gas constant in the corresponding units: [tex]R \approx 8.31\; \rm m^{3}\cdot Pa \cdot mol^{-1} \cdot K^{-1}[/tex].
Let [tex]n[/tex] denote the number of moles of this gas in that [tex]V = 2\; \rm dm^{3}[/tex]. By the ideal gas law, if this gas is an ideal gas, then the following equation would hold:
[tex]P \cdot V = n \cdot R \cdot T[/tex].
Rearrange this equation and solve for [tex]n[/tex]:
[tex]\begin{aligned}n &= \frac{P \cdot V}{R \cdot T} \\ &\approx \frac{1.01325 \times 10^{5}\; {\rm Pa} \times 2 \times 10^{-3}\; {\rm m^{3}}}{8.31 \; {\rm m^{3} \cdot Pa \cdot mol^{-1} \cdot K^{-1}} \times 293.15\; {\rm K}} \\ &\approx 0.08\; \rm mol\end{aligned}[/tex].
In other words, there is approximately [tex]2\; \rm mol[/tex] of this gas in that [tex]V = 2\; \rm dm^{3}[/tex].
