Question

The work done during isothermal irreversible expansion of 2 moles of helium from 2dm³ to 4 dm³ at 1 bar pressure and at 298K is ______.

a) 2.0 kJ
c) 0.2 kJ
b) -2.0 kJ
d) - 0.2 kJ ​

Answer 1

The work done during the isothermal irreversible expansion of 2 moles of helium from 2dm³ to 4dm³ at 1 bar pressure is -0.2 kJ.

Step-by-step explanation:

The work done during an isothermal irreversible expansion of a gas can be calculated by using the formula W = -P_{ext} \Delta V, where W is the work done by the system, P_{ext} is the external pressure, and \(\Delta V\) is the change in volume. For an irreversible process, the external pressure is constant. Given that we have 2 moles of helium gas undergoing an expansion from 2 dm³ to 4 dm³ at 1 bar pressure, the external pressure (P_{ext}) is 1 bar, which is equivalent to 1×10µ N/m². The change in volume \(\Delta V\) is 4 dm³ - 2 dm³; that is 2 dm³ or 2×10³ m³ (since 1 dm³ = 10³ m³).

Calculating the work done:

W = -P_{ext} \Delta V

W = -(1×10µ N/m²) (2×10³ m³)

W = -2×10³ N·m

W = -2×10³ J

Note that 1 J = 1 N·m, and since 1 kJ = 10³ J, we convert the work done to kilojoules:

W = -0.2 kJ

Therefore, the work done during the isothermal irreversible expansion of helium in this scenario is -0.2 kJ.