Question
upstudy study bank question image url

Calcular el volumen molar de H 2 a 350 K y 2.3 atm usando: a) Ecuación de gas ideal. b) Ecuación de gas de Van der Walls. c) Por medio del factor de Compresibilidad. d) ecuación de Peng/Robinson

Ask by Ward Adkins. in Mexico
Mar 18,2025

Upstudy AI Solution

Tutor-Verified Answer

Answer

El volumen molar de \( \text{H}_2 \) a \( 350 \, \text{K} \) y \( 2.3 \, \text{atm} \) es aproximadamente \( 12.49 \, \text{L/mol} \) según la ecuación de gas ideal, \( 12.51 \, \text{L/mol} \) con la ecuación de Van der Waals, y \( 12.49 \, \text{L/mol} \) utilizando el factor de compresibilidad.

Solution

Sign in to Unlock Answers for Free!

A Learning Platform Trusted by Millions of Real Students and Teachers.

star-icon Unlock

Answered by UpStudy AI and reviewed by a Professional Tutor

Extra Insights

Para calcular el volumen molar de H2 utilizando la ecuación de gas ideal \( PV = nRT \), podemos reorganizarla para encontrar \( V_m = \frac{RT}{P} \). Usando la constante de gas 0.0821 L·atm/(K·mol), y sustituyendo \( R \), \( T = 350 \, K \) y \( P = 2.3 \, atm \), obtenemos un volumen molar de aproximadamente 6.73 L/mol. La ecuación de Van der Waals ajusta el comportamiento no ideal de los gases. Para H2, con los valores de \( a = 0.244 L^2·atm/mol^2 \) y \( b = 0.0266 L/mol \), se usa la ecuación modificada \( (P + \frac{a}{V_m^2})(V_m - b) = RT \). Al resolver esta ecuación, obtendremos un volumen molar más bajo que el calculado por la ecuación ideal, reflejando las interacciones entre las moléculas de H2 a esa presión y temperatura. ¿Te gustaría que hablemos sobre cómo calcularlo utilizando el factor de compresibilidad y la ecuación de Peng-Robinson también?

Related Questions

Complete each of the following nuclear decay equations by determining the mass number and atomic number of the following isotopes after they have emitted either a beta particle, an alpha particle or gamma radiation. Use the periodic table to assist you in identifying the remaining particle. a. \( \begin{array}{r}210 \\ 82 \\ \mathrm{~Pb}\end{array} \rightarrow \quad{ }_{-1}^{0} \mathrm{e}+ \) \( \qquad \) 1. \( \begin{array}{lll}\frac{1}{\vdots} & \mathrm{~T} \\ \vdots & \rightarrow & 0 \gamma+ \\ \vdots & & 0\end{array} \) \( \qquad \) b. \( { }_{84}^{209} \mathrm{PO} \rightarrow{ }_{82}^{205} \mathrm{~Pb}+ \) \( \qquad \) j. \( { }_{90}^{234} \mathrm{Th} \rightarrow{ }_{91}^{234} \mathrm{~Pa}+ \) \( \qquad \) c. \( { }_{92}^{239} \mathrm{U} \rightarrow \quad{ }_{-1}^{0} \mathrm{e}+ \) \( \qquad \) d. \( { }_{92}^{238} \mathrm{U} \rightarrow{ }_{2}^{4} \mathrm{He}+ \) \( \qquad \) k. \( { }_{94}^{239} \mathrm{Pu} \rightarrow \quad{ }_{2}^{4} \mathrm{He}+ \) \( \qquad \) I. \( { }_{13}^{27} \mathrm{Al} \rightarrow{ }_{2}^{4} \mathrm{He}+ \) \( \qquad \) e. \( \begin{array}{r}228 \\ 93 \\ \mathrm{~Np}\end{array} \rightarrow \quad-1 \mathrm{e}+ \) \( \qquad \) f. \( \begin{array}{l}42 \\ 19\end{array} \mathrm{~K} \quad{ }_{19}^{42} \mathrm{~K}+ \) \( \qquad \) m. \( { }_{13}^{27} \mathrm{Al} \rightarrow{ }_{-1}^{0} \mathrm{e}+ \) \( \qquad \) g. \( { }_{88}^{226} \mathrm{Ra} \rightarrow{ }_{2}^{4} \mathrm{He}+ \) \( \qquad \) ก. \( { }_{4}^{9} \mathrm{Be} \rightarrow{ }_{4}^{9} \mathrm{Be}+ \) \( \qquad \)

Latest Chemistry Questions

Complete each of the following nuclear decay equations by determining the mass number and atomic number of the following isotopes after they have emitted either a beta particle, an alpha particle or gamma radiation. Use the periodic table to assist you in identifying the remaining particle. a. \( \begin{array}{r}210 \\ 82 \\ \mathrm{~Pb}\end{array} \rightarrow \quad{ }_{-1}^{0} \mathrm{e}+ \) \( \qquad \) 1. \( \begin{array}{lll}\frac{1}{\vdots} & \mathrm{~T} \\ \vdots & \rightarrow & 0 \gamma+ \\ \vdots & & 0\end{array} \) \( \qquad \) b. \( { }_{84}^{209} \mathrm{PO} \rightarrow{ }_{82}^{205} \mathrm{~Pb}+ \) \( \qquad \) j. \( { }_{90}^{234} \mathrm{Th} \rightarrow{ }_{91}^{234} \mathrm{~Pa}+ \) \( \qquad \) c. \( { }_{92}^{239} \mathrm{U} \rightarrow \quad{ }_{-1}^{0} \mathrm{e}+ \) \( \qquad \) d. \( { }_{92}^{238} \mathrm{U} \rightarrow{ }_{2}^{4} \mathrm{He}+ \) \( \qquad \) k. \( { }_{94}^{239} \mathrm{Pu} \rightarrow \quad{ }_{2}^{4} \mathrm{He}+ \) \( \qquad \) I. \( { }_{13}^{27} \mathrm{Al} \rightarrow{ }_{2}^{4} \mathrm{He}+ \) \( \qquad \) e. \( \begin{array}{r}228 \\ 93 \\ \mathrm{~Np}\end{array} \rightarrow \quad-1 \mathrm{e}+ \) \( \qquad \) f. \( \begin{array}{l}42 \\ 19\end{array} \mathrm{~K} \quad{ }_{19}^{42} \mathrm{~K}+ \) \( \qquad \) m. \( { }_{13}^{27} \mathrm{Al} \rightarrow{ }_{-1}^{0} \mathrm{e}+ \) \( \qquad \) g. \( { }_{88}^{226} \mathrm{Ra} \rightarrow{ }_{2}^{4} \mathrm{He}+ \) \( \qquad \) ก. \( { }_{4}^{9} \mathrm{Be} \rightarrow{ }_{4}^{9} \mathrm{Be}+ \) \( \qquad \)
Try Premium now!
Try Premium and ask Thoth AI unlimited math questions now!
Maybe later Go Premium
Study can be a real struggle
Why not UpStudy it?
Select your plan below
Premium

You can enjoy

Start now
  • Step-by-step explanations
  • 24/7 expert live tutors
  • Unlimited number of questions
  • No interruptions
  • Full access to Answer and Solution
  • Full Access to PDF Chat, UpStudy Chat, Browsing Chat
Basic

Totally free but limited

  • Limited Solution
Welcome to UpStudy!
Please sign in to continue the Thoth AI Chat journey
Continue with Email
Or continue with
By clicking “Sign in”, you agree to our Terms of Use & Privacy Policy