Question

With pressure held constant at 210 kPa, what was the change in the internal energy of the gas when it expanded from 0.75 m3 to 1.9 m3?

a. 210 kJ
b. 262.5 kJ
c. 315 kJ
d. 420 kJ

Answer 1

The change in internal energy of the gas cannot be determined without the value of heat transferred.

Step-by-step explanation:

The change in internal energy of a gas can be calculated using the equation:

ΔU = Q - W

Where ΔU is the change in internal energy, Q is the heat transferred to the gas, and W is the work done by the gas.

In this case, the problem states that the pressure is held constant at 210 kPa. Since the pressure is constant, the work done by the gas is zero (W = 0) because the formula for work done in an isobaric process is:

W = P(ΔV)

Since the pressure is constant, the change in volume (ΔV) is the only term that affects the work done. Therefore, with a constant pressure and no change in volume, there is no work done (W = 0). This means that the change in internal energy (ΔU) is equal to the heat transferred to the gas (Q).

Since the problem does not provide the value of Q, we cannot calculate the change in internal energy (AU). Therefore, we cannot determine the answer to this question based on the information given.