A) initial volume
We can calculate the initial volume of the gas by using the ideal gas law:
![p_i V_i = nRT_i](https://tex.z-dn.net/?f=p_i%20V_i%20%3D%20nRT_i)
where
![p_i=1.0 atm=1.01 \cdot 10^5 Pa](https://tex.z-dn.net/?f=p_i%3D1.0%20atm%3D1.01%20%5Ccdot%2010%5E5%20Pa)
is the initial pressure of the gas
![V_i](https://tex.z-dn.net/?f=V_i)
is the initial volume of the gas
![n=2.3 mol](https://tex.z-dn.net/?f=n%3D2.3%20mol)
is the number of moles
![R=8.31 J/K mol](https://tex.z-dn.net/?f=R%3D8.31%20J%2FK%20mol)
is the gas constant
![T_i=240^{\circ}C=513 K](https://tex.z-dn.net/?f=T_i%3D240%5E%7B%5Ccirc%7DC%3D513%20K)
is the initial temperature of the gas
By re-arranging this equation, we can find
![V_i](https://tex.z-dn.net/?f=V_i)
:
![V_i = \frac{nRT_i}{p_i} = \frac{(2.3 mol)(8.31 J/mol K)(513 K)}{1.01 \cdot 10^5 Pa}=0.097 m^3](https://tex.z-dn.net/?f=V_i%20%3D%20%20%5Cfrac%7BnRT_i%7D%7Bp_i%7D%20%3D%20%5Cfrac%7B%282.3%20mol%29%288.31%20J%2Fmol%20K%29%28513%20K%29%7D%7B1.01%20%5Ccdot%2010%5E5%20Pa%7D%3D0.097%20m%5E3%20)
2) Now the gas cools down to a temperature of
![T_f = 14^{\circ}C=287 K](https://tex.z-dn.net/?f=T_f%20%3D%2014%5E%7B%5Ccirc%7DC%3D287%20K)
while the pressure is kept constant:
![p_f = p_i = 1.01 \cdot 10^5 Pa](https://tex.z-dn.net/?f=p_f%20%3D%20p_i%20%3D%201.01%20%5Ccdot%2010%5E5%20Pa)
, so we can use again the ideal gas law to find the new volume of the gas
![V_f = \frac{nRT_f}{p_f}= \frac{(2.3 mol)(8.31 J/molK)(287 K)}{1.01 \cdot 10^5 Pa} = 0.054 m^3](https://tex.z-dn.net/?f=V_f%20%3D%20%20%5Cfrac%7BnRT_f%7D%7Bp_f%7D%3D%20%5Cfrac%7B%282.3%20mol%29%288.31%20J%2FmolK%29%28287%20K%29%7D%7B1.01%20%5Ccdot%2010%5E5%20Pa%7D%20%3D%200.054%20m%5E3)
3) In a process at constant pressure, the work done by the gas is equal to the product between the pressure and the difference of volume:
![W=p \Delta V= p(V_f -V_i)](https://tex.z-dn.net/?f=W%3Dp%20%5CDelta%20V%3D%20p%28V_f%20-V_i%29)
by using the data we found at point 1) and 2), we find
![W=p(V_f -V_i)=(1.01 \cdot 10^5 Pa)(0.054 m^3-0.097 m^3)=-4343 J](https://tex.z-dn.net/?f=W%3Dp%28V_f%20-V_i%29%3D%281.01%20%5Ccdot%2010%5E5%20Pa%29%280.054%20m%5E3-0.097%20m%5E3%29%3D-4343%20J)
where the negative sign means the work is done by the surrounding on the gas.
Given that,
Initial force, F = 12 N
First initial charge, q₁ = 3C
First new charge, q₁' = 6C
Initial distance, r = 15 cm
New distance, r' = 45 cm
To find,
The new force of attraction.
Solution,
The force between two charges is given by :
![F=k\dfrac{q_1q_2}{r^2}](https://tex.z-dn.net/?f=F%3Dk%5Cdfrac%7Bq_1q_2%7D%7Br%5E2%7D)
![12=k\dfrac{3\times q_2}{15^2}\ ....(1)](https://tex.z-dn.net/?f=12%3Dk%5Cdfrac%7B3%5Ctimes%20q_2%7D%7B15%5E2%7D%5C%20....%281%29)
Let F' is the new force.
![F'=k\dfrac{q_1'q_2'}{r'^2}](https://tex.z-dn.net/?f=F%27%3Dk%5Cdfrac%7Bq_1%27q_2%27%7D%7Br%27%5E2%7D)
![F'=\dfrac{k\times 6\times q_2}{(45)^2}\ ...(2)](https://tex.z-dn.net/?f=F%27%3D%5Cdfrac%7Bk%5Ctimes%206%5Ctimes%20q_2%7D%7B%2845%29%5E2%7D%5C%20...%282%29)
As q₂ is same in this case.
Dividing equation (1) and (2) :
![\dfrac{F}{F'}=\dfrac{k\dfrac{3q_2}{15^2}}{\dfrac{k\times 6\times q_2}{45^2}}\\\\\dfrac{12}{F'}=4.5\\\\F'=\dfrac{12}{4.5}\\\\F'=2.67\ N](https://tex.z-dn.net/?f=%5Cdfrac%7BF%7D%7BF%27%7D%3D%5Cdfrac%7Bk%5Cdfrac%7B3q_2%7D%7B15%5E2%7D%7D%7B%5Cdfrac%7Bk%5Ctimes%206%5Ctimes%20q_2%7D%7B45%5E2%7D%7D%5C%5C%5C%5C%5Cdfrac%7B12%7D%7BF%27%7D%3D4.5%5C%5C%5C%5CF%27%3D%5Cdfrac%7B12%7D%7B4.5%7D%5C%5C%5C%5CF%27%3D2.67%5C%20N)
So, the new force of attraction is 2.67 N.
Frequency. This refers to how often you exercise. The point is to meet your goals without overtraining the body. When it comes to cardio: As a general rule of thumb, aim for a minimum of three cardio sessions per week.