A triply charged ion with velocity 7.00 × 10^6 m/s moves in a path of radius 36.0 cm in a magnetic field of 0.55 T in a mass spe
1 answer:
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Answer:
A. The work done during the process is W = 0
B. The value of heat transfer during the process Q = 442.83
Explanation:
Given Data
Initial pressure = 100 k pa
Initial temperature = 25 degree Celsius = 298 Kelvin
Final pressure = 300 k pa
Vessel is rigid so change in volume of the gas is zero. so that initial volume is equal to final volume.
⇒ =
------------- (1)
Since volume of the gas is constant so pressure of the gas is directly proportional to the temperature of the gas.
⇒ P ∝ T
⇒ =
⇒ Put all the values in the above formula we get the final temperature
⇒ =
× 298
⇒ = 894 Kelvin
(A). Work done during the process is given by W = P × ()
From equation (1), =
so work done W = P × 0 = 0
⇒ W = 0
Therefore the work done during the process is zero.
Heat transfer during the process is given by the formula Q = m (
-
)
Where m = mass of the gas = 1 kg
= specific heat at constant volume of nitrogen = 0.743
Thus the heat transfer Q = 1 × 0.743 × ( 894- 298 )
⇒ Q = 442.83
Therefore the value of heat transfer during the process Q = 442.83