Question

One mole of an ideal gas at 300 K is compressed isothermally from a volume of \(V_1\) to \(V_2\). Calculate:
The work done on the gas
The change in internal energy
The heat exchanged with the surroundings
Use \(R = 8.314\, \text{J/molK}\), \( \ln(2.5) = 0.916\)

Hint:

In an isothermal process, \(\Delta U = 0\), and the heat exchanged equals the work done.

Answer 1

(a)\[ W = -nRT \ln \left(\frac{V_2}{V_1}\right) \]

Where:
- \(n = 1 \, \text{mol}\) (moles of gas)
- \(R = 8.314\, \text{J/molK}\) (ideal gas constant)
- \(T = 300\, \text{K}\) (temperature)
- \(\ln(2.5) = 0.916\)

Thus, work done is:

\[ W = -1 \times 8.314 \times 300 \times 0.916 = -2276.44 \, \text{J} \]

Answer: \(\boxed{-2276.44\, \text{J}}\)

(b) The change in internal energy:
For an ideal gas undergoing an isothermal process, the change in internal energy (\(\Delta U\)) is zero because internal energy of an ideal gas depends only on temperature, and the temperature does not change in an isothermal process.

\[ \Delta U = 0 \]

Answer: \(\boxed{0 \, \text{J}}\)

(c) The heat exchanged with the surroundings:
According to the first law of thermodynamics:

\[ \Delta U = Q - W \]

Since \(\Delta U = 0\), we have:

\[ Q = W = -2276.44 \, \text{J} \]

Answer: \(\boxed{-2276.44 \, \text{J}}\)