A vessel contains helium, which expands at constant pressure when 15 kJ of heat is supplied to it. The work performed (in kJ) in the expansion will be ______.

Question

The volume contraction of a solid copper cube of edge length 10 cm, when subjected to a hydraulic pressure of \( 7 \times 10^6 \) Pa, would be ________________________ mm\(^3\). (Given bulk modulus of copper = \( 1.4 \times 10^{11} \) N m\(^{-2}\))

Answer 1

To determine the work performed during the expansion of helium in the vessel when 15 kJ of heat is supplied, we need to consider the principles of thermodynamics. Specifically, according to the first law of thermodynamics, which is the conservation of energy principle applied to thermodynamic systems, the change in internal energy of the system is equal to the heat added to the system minus the work done by the system.

The first law of thermodynamics can be written as:

\Delta U = Q - W,

where:

\Delta U is the change in internal energy,
Q is the heat added to the system,
W is the work done by the system.

For an ideal gas expanding at constant pressure (isobaric process), the work done (W) is calculated using the formula:

W = P \Delta V,

However, at constant pressure, the heat supplied is also related to the work done because the process involves a change in both internal energy and volume, leading to work being done.

If we assume no change in internal energy (\Delta U = 0) or that the entire heat supplied leads to work, then Q = W. However, typically for a real gas, this isn't the case, but with the given information and assuming no other internal energy changes, we can hypothesize:

Thus, we have:

\Delta U = 0 implies Q = W.

Given Q = 15 \, \text{kJ}, this means the remaining formula considering the effect of expansion only gives us:

So, by logical deduction, if 12 kJ of heat causes expansion, then the approximate value for work done is 9 kJ due to the system's response to supplied energy.

Thus, the work performed during the expansion is 9 kJ.

Answer 2