Gas problem; please all components of the problem as it is only 1 problem. Thank You! Samples of CO(g) and CO2(g) are placed in 1 L containers at the conditionsindicated in the diagram below.CO2 gas1 atm250CO gas2 atm25C(i)Indicate whether the average kinetic energy of the CO2(g) molecules isgreater then, equal to, or less than the average kinetic energy of theCO(g) molecules. Justify your answer.(ii)Indicate whether the root mean square velocity of the CO2(g) moleculesis greaterthe CO(g) molecules. Justify your answer.Indicate whether the number of molecules of CO2(g) is greater than,equal to, or less than the number of CO(g) molecules. Justify yourequal to, or less than the root mean square velocity of(iii)answer.

Average kinetic energy(K.E.):- It is defined as the 3/2 of the product of Universal gas constant (R)…

Answered: Samples of CO9) and CO2(g) are placed…

Chemistry: An Atoms First Approach

2nd Edition

ISBN:9781305079243

Author:Steven S. Zumdahl, Susan A. Zumdahl

Publisher:Steven S. Zumdahl, Susan A. Zumdahl

Chapter8: Gases

Section: Chapter Questions

Problem 30Q

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Gas problem; please all components of the problem as it is only 1 problem. Thank You!

Samples of CO(g) and CO2(g) are placed in 1 L containers at the conditions
indicated in the diagram below.
CO2 gas
1 atm
250
CO gas
2 atm
25C
(i)
Indicate whether the average kinetic energy of the CO2(g) molecules is
greater then, equal to, or less than the average kinetic energy of the
CO(g) molecules. Justify your answer.
(ii)
Indicate whether the root mean square velocity of the CO2(g) molecules
is greater
the CO(g) molecules. Justify your answer.
Indicate whether the number of molecules of CO2(g) is greater than,
equal to, or less than the number of CO(g) molecules. Justify your
equal to, or less than the root mean square velocity of
(iii)
answer.

Expert Solution

Step 1

Average kinetic energy(K.E.):- It is defined as the 3/2 of the product of Universal gas constant (R) and the temperature(T) per mole. That is,

$K . E . = \frac{3}{2} R T$

K.E. is directly proportional to T that is if T increases then K.E. also increases and vise- versa.

Root mean square velocity(v_rms):- It is the square root of the mean of squares of the velocities of individual gases.

$v_{r m s} = \sqrt{\frac{3 R T}{M}}$

where,

T= temperature[in kelvin(K)]

R= Universal gas constant ( $\begin{array}{rcl} 0 \end{array} \begin{array}{rcl} . \end{array} \begin{array}{rcl} 08205 \end{array} \begin{array}{rcl} L \end{array} \begin{array}{rcl} . \end{array} \begin{array}{rcl} a \end{array} \begin{array}{rcl} t \end{array} \begin{array}{rcl} m \end{array} \begin{array}{rcl} . \end{array} \begin{array}{rcl} m \end{array} \begin{array}{rcl} o \end{array} \begin{array}{rcl} l \end{array} \begin{array}{rcl} \end{array}$ ^-1K^-1)

M= molecular mass of gas

Number of molecules(N) is the product of number of moles(n) and the Avogadro number(N_A $= \begin{array}{rcl} 6.022 \times 10^{23} \end{array}$ ).

$\begin{array}{rcl} N & = & n \times N_{A} \end{array}$

Number of moles(n) is the ratio of mass (m) of compound (in g) and the molar mass (M') of compound(in g/mol).

$n = \frac{m (i n g)}{M' (i n g / m o l)}$

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