We know, the ideal gas equation,
P1V1 / T1 = P2V2 / T2
Here, P1 = 760 mm
V1 = 10 m3
T1 = 27 + 273 = 300 K
P2 = 400 mm Hg
T2 = -23 + 273 = 250 K
Substitute their values,
760*10 / 300 = 400 * V2 / 250
25.33 * 250 = 400 * V2
V2 = 6333.333/ 400
V2 = 15.83
In short, Your Answer would be approx. 15.83 m3
Hope this helps!
Answer:
The forces creating the net force must lie in the same direction.
Explanation:
newton's second law states that the net force acting on the body is equal to the product of mass and the acceleration of the body.
If there are several forces acting on the body in different directions, then we have to find teh net force by using the vector sum and then find the acceleration.
It is not necessary that all the forces acting in the same direction.
if they are in different directions then we have to find the net force by t=using the formula for the vector sum.
Answer:0.061
Explanation:
Given
Temperature of soup 
heat capacity of soup 
Here Temperature of soup is constantly decreasing
suppose T is the temperature of soup at any instant
efficiency is given by
integrating From 
to 
![W=c_v\left [ T-T_C\ln T\right ]_{T_H}^{T_C}](https://tex.z-dn.net/?f=W%3Dc_v%5Cleft%20%5B%20T-T_C%5Cln%20T%5Cright%20%5D_%7BT_H%7D%5E%7BT_C%7D)
![W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]](https://tex.z-dn.net/?f=W%3Dc_v%5Cleft%20%5B%20%5Cleft%20%28%20T_C-T_H%5Cright%20%29-T_C%5Cleft%20%28%20%5Cln%20%5Cfrac%7BT_C%7D%7BT_H%7D%5Cright%20%29%5Cright%20%5D)
Now heat lost by soup is given by
Fraction of the total heat that is lost by the soup can be turned is given by
![=\frac{c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]}{c_v(T_C-T_H)}](https://tex.z-dn.net/?f=%3D%5Cfrac%7Bc_v%5Cleft%20%5B%20%5Cleft%20%28%20T_C-T_H%5Cright%20%29-T_C%5Cleft%20%28%20%5Cln%20%5Cfrac%7BT_C%7D%7BT_H%7D%5Cright%20%29%5Cright%20%5D%7D%7Bc_v%28T_C-T_H%29%7D)
