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.
T = ??°K, V = 15.50L, P = 870torr
PV = nRT, with P in atm, 760torr/atm
T = PV / nR = 1atm/760torr × 870torr×15.5 ÷ .0821 = 216.12 °K
Answer:
W=Fd
This is the equation for it
Answer:
Explanation:
The period of oscillation is given as
T=2π√m/k
Making k subject of the formula
Square both sides of the equation
T²=4π²(m/k)
Cross multiply
T²k=4π²m
Then, divide through by T²
k=4π²m/T²
Where
k is spring constant
m is the mass of the bob
And T is the period of the oscillation
m=140g=0.14kg
14 oscillations takes 14 seconds
Then the period is
T=time/oscillation
T=14/14
T=1sec
Then,
k=4π²m/T²
k=4π²×0.14/1²
k=1.76N/m
Then, the spring constant is 1.76N/m