For an isothermal process, the work done by or on a system of ideal gas is equal to the change in what?
1 answer:
Explanation:
The equation for work done by a system is as follows.
where, = change in internal energy
Q = heat absorbed or released by the system
W = work done by the system
Since it is given that it is an isothermal process. Therefore, .
Whereas internal energy depends on temperature and as there is no change in temperature so, there will be no change in internal energy of the system.
Hence, the equation will be as follows.
0 = Q - W
Q = W
Therefore, we can conclude that for an isothermal process, the work done by or on a system of ideal gas is equal to the change in Q.
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Answer:
Speed of both blocks after collision is 2 m/s
Explanation:
It is given that,
Mass of both blocks, m₁ = m₂ = 1 kg
Velocity of first block, u₁ = 3 m/s
Velocity of other block, u₂ = 1 m/s
Since, both blocks stick after collision. So, it is a case of inelastic collision. The momentum remains conserved while the kinetic energy energy gets reduced after the collision. Let v is the common velocity of both blocks. Using the conservation of momentum as :
v = 2 m/s
Hence, their speed after collision is 2 m/s.