There are some missing data in the text of the exercise. Here the complete text:
"<span>A sample of 20.0 moles of a monatomic ideal gas (γ = 1.67) undergoes an adiabatic process. The initial pressure is 400kPa and the initial temperature is 450K. The final temperature of the gas is 320K. What is the final volume of the gas? Let the ideal-gas constant R = 8.314 J/(mol • K). "
Solution:
First, we can find the initial volume of the gas, by using the ideal gas law:
</span>

<span>where
p is the pressure
V the volume
n the number of moles
R the gas constant
T the absolute temperature
Using the initial data of the gas, we can find its initial volume:
</span>

<span>
Then the gas undergoes an adiabatic process. For an adiabatic transformation, the following relationship between volume and temperature can be used:
</span>

<span>where </span>

for a monoatomic gas as in this exercise. The previous relationship can be also written as

where i labels the initial conditions and f the final conditions. Re-arranging the equation and using the data of the problem, we can find the final volume of the gas:
![V_f = V_i \sqrt[\gamma-1]{ \frac{T_i}{T_f} }=(0.187 m^3) \sqrt[0.67]{ \frac{450 K}{320 K} }=0.310 m^3 = 310 L](https://tex.z-dn.net/?f=V_f%20%3D%20V_i%20%20%5Csqrt%5B%5Cgamma-1%5D%7B%20%5Cfrac%7BT_i%7D%7BT_f%7D%20%7D%3D%280.187%20m%5E3%29%20%5Csqrt%5B0.67%5D%7B%20%5Cfrac%7B450%20K%7D%7B320%20K%7D%20%7D%3D0.310%20m%5E3%20%3D%20310%20L%20%20)
So, the final volume of the gas is 310 L.
Answer:
Landed before it explodes
Explanation:
vf = vi + at,
0 = 145 - (9.8)t,
t = 14.79 s (Time to reach highest point)
14.79 x 2 = 29.59 s (Time to land on the ground)
It will have landed before it explodes because both the time to reach the highest point and the time to land on the ground are less than 32 seconds.
<u>Answers</u>
same distance from the mirror as the object
same size
lateral inversion
<u>Explanation</u>
The image formed by a plane mirror is formed at equal distance from the mirror as the object is.
The size of the image is the same as the size of the object.
The image is laterally inverted. This means that, when a person lifts his right hand, his or her image will raise it's left hand.
Answer:
1.) 1620 km/h^2
2.) 2.7 km
Explanation:
1.) Given that the car start from rest. The initial velocity U will be equal to zero. That is,
U = 0.
Final velocity V = 54 km/h
Time t = 2 minute = 2/60 = 1/30 hour
Acceleration a will be change in velocity per time taken. That is,
a = ( V - U )/ t
Substitute V, U and t into the formula
a = 54 ÷ 1/30
a = 54 × 30 = 1620 km/h ^2
2.) Distance travelled S by the car during the time can be calculated by using the 2nd equation of motion.
S = Ut + 1/2at^2
Substitute all the parameters into the formula
S = 54 × 1/30 + 1/2 × 1620 × (1/30)^2
S = 54/30 + 810 × 1/900
S = 54/30 + 810/900
S = (1620+810)/900
S = 2430/900
S = 2.7 km.
Therefore, distance travelled by the car during this time is 2.7 km
Answer:
1D
2C
3C
4C
5D
6D
7D
8D
9B
Explanation:
better give me points X﹏X