Q3 PE - College Physics (Urone)

Question

(a) Repeat Exercise \({\rm{31}}{\rm{.2}}\), and convert the energy to joules or calories. (b) If all of this energy is converted to thermal energy in the gas, what is its temperature increase, assuming \(50.0\,{\rm{c}}{{\rm{m}}^{\rm{3}}}\) of ideal gas at \({\rm{0}}{\rm{.250 - }}\)atm pressure? (The small answer is consistent with the fact that the energy is large on a quantum mechanical scale but small on a macroscopic scale.)

Step-by-Step Solution

Answer

1. Energy converted in Joule is obtained as: \(1.92 \times {10^{ - 14}}\,{\rm{J}}\).
2. The temperature is obtained as: \(3.03 \times {10^{ - 12}}\,{\rm{K}}\).

Step 1: Define Radioactivity

The spontaneous emission of radiation in the form of particles or high-energy photons as a result of a nuclear process is known as radioactivity.

Step 2: Evaluating the energy in joules or calories

a) Number of ions we have is: \(n = 4 \times {10^3}\,{\rm{ion}}\)

 

Energy required to create one pair of ion is given as:

\({E_{rad}} = 30.0\,{\rm{eV}}\)

The total energy required is then obtained as:

\(\begin{align}{E_{tot}} &= 4000 \times (30.0\,eV)\\ &= 120000\,eV\\ &= 120000 \times 1.6 \times {10^{ - 19}}\,J\\ &= 1.92 \times {10^{ - 14}}\,J\end{align}\)

Therefore, the energy in joule is: \(1.92 \times {10^{ - 14}}\,{\rm{J}}\).

Step 3 : Evaluating the temperature

b) The pressure of the gas is given as:

\(\begin{align}P &= 0.250\,atm{\rm{ }}\\ &= 0.250 \times 101325\,Pa\\P &= 25331\,Pa\end{align}\)

The volume of the gas is given as:

\(\begin{align}V &= 50\,c{m^3}{\rm{ }}\\ &= 5.0x{10^{ - 5}}\,{m^3}\end{align}\)

We the know that:

\(\begin{align}du &= n{c_V}dT\\dT &= \frac{{du}}{{n{c_v}}}\end{align}\)

The equation, for ideal gas is:

\(\begin{align}PV &= nRT\\R &= \frac{{PV}}{{nT}}\end{align}\)

We also know that, for ideal gas we have is:

\(\begin{align}{c_v} &= \frac{3}{2}{\rm{ }}\\R &= \frac{3}{2}\frac{{PV}}{{nT}}\end{align}\)

The experiment with the value of gm tube must be going on at room temperature, So we have the value as: \({\rm{T   =  300 K}}\).

We, then obtain:

\(\begin{align}dT{\rm{ }} &= \frac{{dU}}{{n\frac{3}{2}\frac{{PV}}{{nT}}}}\\ &= \frac{{2dUT}}{{3PV}}\end{align}\)

 

Substitute all the value in the above equation.

\(\begin{align}dT &= \frac{{2\left( {1.92 \times {{10}^{ - 14}}\,J} \right)(300\,K)}}{{3\left( {5.0 \times {{10}^{ - 5}}\,{m^3}} \right)(25331\,Pa)}}\\ &= 3.03 \times {10^{ - 12}}\,K\end{align}\)

 

Therefore, the temperature is: \(3.03 \times {10^{ - 12}}\,{\rm{K}}\).