Respuesta :
Since the process occurs at constant pressure, the work done by the gas on the piston is given by
[tex]W=p \Delta V = p (V_f - V_i)[/tex]
where
W is the work
p is the pressure
[tex]V_f[/tex] is the final volume
[tex]V_i[/tex] is the initial volume
In our problem, we have
[tex]V_i =125 L = 125 \cdot 10^{-3}m^3[/tex]
[tex]V_f = 90 L = 90 \cdot 10^{-3} m^3[/tex]
[tex]W=-104 J[/tex] (where the negative sign means the work is done by the piston on the gas)
We can re-arrange the previous equation and use these values to find the pressure:
[tex]p= \frac{W}{V_f - V_i}= \frac{-104 J}{90 \cdot 10^{-3} m^3 - 125 \cdot 10^{-3} m^3}=2971 Pa [/tex]
[tex]W=p \Delta V = p (V_f - V_i)[/tex]
where
W is the work
p is the pressure
[tex]V_f[/tex] is the final volume
[tex]V_i[/tex] is the initial volume
In our problem, we have
[tex]V_i =125 L = 125 \cdot 10^{-3}m^3[/tex]
[tex]V_f = 90 L = 90 \cdot 10^{-3} m^3[/tex]
[tex]W=-104 J[/tex] (where the negative sign means the work is done by the piston on the gas)
We can re-arrange the previous equation and use these values to find the pressure:
[tex]p= \frac{W}{V_f - V_i}= \frac{-104 J}{90 \cdot 10^{-3} m^3 - 125 \cdot 10^{-3} m^3}=2971 Pa [/tex]
Answer: 3 x 10^5 pascals
Explanation: I think you meant to put 10^4 joules instead of 104, which is why nobody likes the previous answer, using 10^4 joules gets you. the answer with the following:
Using the same equation the person above me posted....
-10^4/(90E-3-125E-3) And then only one significant figure from the question and we round to 3 x 10^5 pascals