An ideal monatomic gas at 300 k expands adiabatically and reversibly to twice its volume. what is its final temperature?
1 answer:
In an adiabatic process, the following relationship holds:
where T is the gas temperature, V is the volume and
is the adiabatic index, which is equal to 
for a monoatomic gas.
We can re-write the equation as
where the labels 1,2 refer to the initial and final conditions of the gas.
Let's rewrite it for
, the final temperature:
We can now substitute the initial temperature, T1=300 K, and
, because the final volume is twice the initial one. So we find the value of the final temperature:
where T is the gas temperature, V is the volume and
We can re-write the equation as
where the labels 1,2 refer to the initial and final conditions of the gas.
Let's rewrite it for
We can now substitute the initial temperature, T1=300 K, and
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Answer:
250 m
Explanation:
The car in this problem is moving of uniform accelerated motion, so we can use the following suvat equation:
where
s is the distance covered
u is the initial velocity
t is the time
a is the acceleration
Assuming the car starts from rest,
u = 0
Also we know that
a = 5 m/s^2 (acceleration of the car)
t = 10 s
Substituting, we find the distance covered: