Answer: The work done in J is 324
Explanation:
To calculate the amount of work done for an isothermal process is given by the equation:
![W=-P\Delta V=-P(V_2-V_1)](https://tex.z-dn.net/?f=W%3D-P%5CDelta%20V%3D-P%28V_2-V_1%29)
W = amount of work done = ?
P = pressure = 732 torr = 0.96 atm (760torr =1atm)
= initial volume = 5.68 L
= final volume = 2.35 L
Putting values in above equation, we get:
![W=-0.96atm\times (2.35-5.68)L=3.20L.atm](https://tex.z-dn.net/?f=W%3D-0.96atm%5Ctimes%20%282.35-5.68%29L%3D3.20L.atm)
To convert this into joules, we use the conversion factor:
![1L.atm=101.33J](https://tex.z-dn.net/?f=1L.atm%3D101.33J)
So, ![3.20L.atm=3.20\times 101.3=324J](https://tex.z-dn.net/?f=3.20L.atm%3D3.20%5Ctimes%20101.3%3D324J)
The positive sign indicates the work is done on the system
Hence, the work done for the given process is 324 J
Answer: 8400 J
Explanation:
The formula referenced in the question is:
Where:
is the thermal energy
is the mass of the water sample
is the specific heat capacity of water
is the variation in temperature
Solving:
This is the thermal energy released
Answer:
F = 3.98 kN
Explanation:
GIVEN DATA:
sides of box = 17 cm
pressure = 1 atm = 101325 N/m2
T2 = 378K
T1 = 278 K
final pressure can be calculate by using below relation
![\frac{P_{1}}{P_{2}}=\frac{T_{1}}{T_{2}}](https://tex.z-dn.net/?f=%5Cfrac%7BP_%7B1%7D%7D%7BP_%7B2%7D%7D%3D%5Cfrac%7BT_%7B1%7D%7D%7BT_%7B2%7D%7D)
we know that
force = pressure * area
therefore force is
![F =(\frac{T_{1}}{T_{2}}*P_{1})A](https://tex.z-dn.net/?f=F%20%3D%28%5Cfrac%7BT_%7B1%7D%7D%7BT_%7B2%7D%7D%2AP_%7B1%7D%29A)
![F =(\frac{378}{278}*101325)(17*10^{-2})^{2}](https://tex.z-dn.net/?f=F%20%3D%28%5Cfrac%7B378%7D%7B278%7D%2A101325%29%2817%2A10%5E%7B-2%7D%29%5E%7B2%7D)
F = 3.98 kN
You know you can skip those and just submit them, they don’t even check them