During an isothermal compression of an ideal gas, 410J of heat must be removed from the gas to maintain constant temperature. How much work is done by the gas during the process?

Work done = -410J

For an isothermal process, temperature is constant and thus the heat transfer into and out of the system happens slowly so that the system is in thermal equilibrium.

Also, from the first law of thermodynamics, the change in internal energy (ΔU) of a closed system is equal to the sum of the heat transferred (Q) and the work done (W) by the system. i.e

ΔU = W + Q         (i)

The internal energy is dependent on the temperature of the system and since this temperature is constant the change in internal energy is zero (0).

i.e

ΔU = 0

Substitute this value into equation (i) as follows;

⇒ 0 = W + Q

⇒ W = -Q              (ii)

Therefore, for such process, the work done is the negative of the heat transferred (heat removed in this case).

Given

Q = 410J

Substitute into equation (ii)

W = -410J

Therefore, the work done is -410 J

The value is negative to show that the process is a compression meaning that work is done by the surrounding on the gas.

answer: i pick a

explanation:

the speed of the cart after 3 seconds of low fan speed is ⇒ 54.0 cm/s.

the speed of the cart after 5 seconds of medium fan speed is ⇒ 120.0 cm/s.

the speed of the cart after 2 seconds of high fan speed is ⇒ 64.0 cm/s. 

just did the lab.