Q1.E - University Physics with Modern Physics

Question

A 20.0L tank contains  of helium at . The molar mass of helium is 4.00 g/mol. (a) How many moles of helium are in the tank? (b) What is the pressure in the tank, in pascals and in atmospheres?

Step-by-Step Solution

Answer

(a)There are 0.122moles of Helium in the tank.

(b)The pressure in the tank in Pascals is , and in atmospheres, it is .

Step 1: Relations for number of moles and pressure

The number of moles n is the ratio of mass to molecular mass, and is expressed as, $n=\frac{m}{M}$where n is the number of moles, m is the mass of the sample taken, and M is the molecular mass.

 

Pressure is calculated from the ideal gas equation PV=nRT  where P is the pressure, V , the volume, T the temperature, n the number of moles.

Rearranging, the expression of pressure is obtained as $P=\frac{nRT}{V}$ ………………………(1)

Step 2: Calculation of number of moles

Given the values of mass  m as   $\mathbf{4}\mathbf{.}\mathbf{86}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{4}\mathbf{ }}\mathbf{kg}$ and molar mass M as $4.00\mathrm{g}/\mathrm{mol}=4.00*{10}^{-3}\mathrm{kg}/\mathrm{mol}$.

Substitute the values of  m and M in the expression for number of moles $n=\frac{m}{M}$,

   $$\begin{gathered} n=\frac{m}{M} \\ n=\frac{4.86*{10}^{-4}}{4*{10}^{-3}} \\ n=0.1215moles \end{gathered}$$

Hence, the number of moles is 0.1215.

Step 3: Calculation of pressure

Given the values of number of moles n as 0.1215, temperature T as $18.0°C$ which is 18+273=291 , volume V as  20L = $20*{10}^{-3} m^3$

The universal gas constant  $\mathrm{R}=8.314\mathrm{J}\times {\mathrm{mol}}^{-1}\times {\mathrm{K}}^{-1}$.

Substituting the values of n, R, T, V in the equation for pressure $P=\frac{nRT}{V}$

                                                                 $$\begin{gathered} P = \frac{nRT}{V} \\ P=\frac{0.1215*8.314*291}{20*{10}^{-3}} \\ P=14697 \\ P=1.4697*{10}^{4}Pa \end{gathered}$$

Thus, the value of pressure in Pascals is $1.4697*{10}^{4}Pa$.

 

Convert the value in pascals into that in atmosphere by the relation $1atm=1.013*{10}^{5}Pa$

                                                  Thus,$$\begin{gathered} P_{atm}=\frac{P_{Pa}}{1.013} \\ {P}_{atm}=\frac{1.4697*{10}^4}{1.013*{10}^5} \\ {P}_{a}tm=0.1451atm \end{gathered}$$             

 

Hence, the value of pressure in atmosphere is 0.1451atm.