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nitrogen gas at standard atmospheric pressure 101.3kPa has a volume of 0.080m^3 . if there are 3.0 mol of gas what is the temperature of nitrogen gas

Sagot :

jacob193

Answer:

Approximately [tex]52\; {\rm ^\circ C}[/tex] (approximately [tex]325\; \rm K[/tex]), assuming that nitrogen is an ideal gas.

Explanation:

  • Let [tex]P[/tex] denote the pressure of this nitrogen gas sample.
  • Let [tex]V[/tex] denote the volume of this nitrogen gas sample.
  • Let [tex]n[/tex] denote the number of moles of [tex]\rm N_{2}[/tex] molecules in this nitrogen gas sample.
  • Let [tex]T[/tex] denote the absolute temperature of this nitrogen gas sample (typically measured in degrees kelvins.)

Let [tex]R[/tex] denote the ideal gas constant. By the ideal gas law, the following equation would relate these quantities:

[tex]P \cdot V = n \cdot R \cdot T[/tex].

Rearrange this equation to obtain an expression for [tex]T[/tex]:

[tex]\begin{aligned}T &= \frac{P \cdot V}{n \cdot R}\end{aligned}[/tex].

Look up the ideal gas constant: [tex]R \approx 8.314\; \rm Pa \cdot m^{3} \cdot K^{-1} \cdot mol^{-1}[/tex].

Convert each measurements from the question to standard units:

  • [tex]P = 101.3\; \rm kPa = 101.3 \times 10^{3}\; \rm Pa[/tex].
  • [tex]V = 0.080\; \rm m^{3}[/tex].
  • [tex]n = 3.0\; \rm mol[/tex].

Substitute these values into the expression for [tex]T[/tex]:

[tex]\begin{aligned}T &= \frac{P \cdot V}{n \cdot R} \\ &\approx \frac{101.3\times 10^{5}\; \rm Pa \times 0.080\; \rm m^{3}}{3.0\; \rm mol \times 8.314\; \rm Pa \cdot m^{3} \cdot K^{-1} \cdot mol^{-1}} \\ &\approx 324.91\; \rm K\end{aligned}[/tex].

Convert the unit of this temperature to degrees celsius:

[tex]\begin{aligned} & 324.91\; \rm K \\ =\; & (324.91 - 273.15)\; {\rm ^\circ C} \\ \approx \; & 52\; {\rm ^\circ C} \end{aligned}[/tex].

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