Calculating Number Density of Gas Molecules

[Chase11]

Homework Statement

What is the number density, N/V, of gas molecules in the closed vessel in the previous problem?

(Previous problem: A closed vessel contains an ideal gas at an absolute pressure of 1.5 atm and a termperature of 42 degrees celsius. To what final temperature (in °C) must the gas be heated so its final pressure is 2.0 atm?

Homework Equations

pV=N$k\ =\ 1.3806503(24)\ \times\ 10^{-23}\ J\ K^{-1}$T

The Attempt at a Solution

From the previous problem I have p=2.0 atm, T=420K and reorganizing the equation I get N/V=p/$k\ =\ 1.3806503(24)\ \times\ 10^{-23}\ J\ K^{-1}$T

so N/V=2.0atm/$k\ =\ 1.3806503(24)\ \times\ 10^{-23}\ J\ K^{-1}$(420K) = 3.45x10^20. The answer is supposed to be 3.5x10^25. I cannot figure out why I am getting the wrong answer, can anybody shed some light on this?

 

[SteamKing]

Have you checked the units of the quantities you have plugged into your formula?

 

[Chestermiller]

Homework Statement

What is the number density, N/V, of gas molecules in the closed vessel in the previous problem?

(Previous problem: A closed vessel contains an ideal gas at an absolute pressure of 1.5 atm and a termperature of 42 degrees celsius. To what final temperature (in °C) must the gas be heated so its final pressure is 2.0 atm?

Homework Equations

pV=N$k\ =\ 1.3806503(24)\ \times\ 10^{-23}\ J\ K^{-1}$T

The Attempt at a Solution

From the previous problem I have p=2.0 atm, T=420K and reorganizing the equation I get N/V=p/$k\ =\ 1.3806503(24)\ \times\ 10^{-23}\ J\ K^{-1}$T

so N/V=2.0atm/$k\ =\ 1.3806503(24)\ \times\ 10^{-23}\ J\ K^{-1}$(420K) = 3.45x10^20. The answer is supposed to be 3.5x10^25. I cannot figure out why I am getting the wrong answer, can anybody shed some light on this?

At the initial conditions, I get a molar density of 0.0581 moles/liter. Multiplying this by avagodro's number, I get 3.5 x 10²² molecules/liter. Per cubic meter, this is 3.5 x 10²⁵ molecules. The same result is obtained at the final conditions.

Chet