Skip to content
Problems · Problem 5.2

Q.Two litres of an ideal gas at a pressure of 10 atm expands isothermally at 25 °C into a vacuum until its total volume is 10 litres. How much heat is absorbed and how much work is done in the expansion?

Punjab PsebTextbookSubjective· 2mImportance★★★★★est
2% · 2/98 Questions
✓ Free question

For an isothermal free expansion into vacuum, no work is done (W=0W = 0) and no heat is exchanged (Q=0Q = 0) because the gas does not push against any external pressure and the internal energy of an ideal gas depends only on temperature.

The question describes a classic scenario: an ideal gas expanding into a vacuum. The key is to recognise that "expands into a vacuum" means the external pressure is zero. This is a free expansion.

Let’s start with the First Law of Thermodynamics:

ΔU=Q+W\Delta U = Q + W

Here, QQ is heat absorbed by the system, WW is work done on the system (many textbooks use WW as work done by the system, so the sign convention matters — we’ll stick with the physics convention where WW is work done on the gas; if work is done by the gas, WW is negative).

For an ideal gas, internal energy UU depends only on temperature. Since the process is isothermal (temperature constant at 25 °C), the change in internal energy is zero:

ΔU=0\Delta U = 0

Now, work done during expansion against an external pressure PextP_{\text{ext}} is:

W=−∫Pext dVW = - \int P_{\text{ext}} \, dV

The negative sign appears because when the gas expands (dV>0dV > 0), work is done by the gas, so work done on the gas is negative.

In a free expansion into vacuum, Pext=0P_{\text{ext}} = 0. Therefore:

W=−∫(0) dV=0W = - \int (0) \, dV = 0

From the First Law:

0=Q+0⇒Q=00 = Q + 0 \quad \Rightarrow \quad Q = 0

So no heat is absorbed and no work is done.

Watch out

A common mistake is to try to calculate work using W=−nRTln⁡(V2/V1)W = -nRT \ln(V_2/V_1) for an isothermal process. That formula applies only when the expansion is reversible — i.e., when the external pressure is infinitesimally less than the gas pressure at every step. In a free expansion, the external pressure is zero throughout, so no work is done. The reversible formula gives the maximum work possible, not the actual work here.

Tip

Think of it this way: if you push against a wall that isn’t there, you do no work. The gas expands into empty space — there’s nothing to push against. So no energy is transferred as work, and since temperature doesn’t change, no heat flows either.

✓Final answer

The heat absorbed is 00 and the work done is 00.

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.