Find answers to your questions and expand your knowledge with IDNLearn.com. Discover the information you need from our experienced professionals who provide accurate and reliable answers to all your questions.
Sagot :
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].
Thank you for using this platform to share and learn. Don't hesitate to keep asking and answering. We value every contribution you make. For clear and precise answers, choose IDNLearn.com. Thanks for stopping by, and come back soon for more valuable insights.