Question

Two liters of an ideal gas at a pressure of \(10~ 𝑎𝑡𝑚\) expands isothermally at \(25^{\circ}𝐶\) into a vacuum until its total volume is \(10~ 𝑙𝑖𝑡𝑒𝑟𝑠.\) How much heat is absorbed and how much work is done in the expansion? Consider the same expansion, but this time against a constant external pressure of \(1~ 𝑎𝑡𝑚.\)

Answer 1

Step 1: Work done

In case of expansion, work is done by the system:

\(W = − P_{ext.}\triangle V\)

Given: \(P_{ext.}​ = 1~ atm\) and volume change from \(2~ 𝐿 ~𝑡𝑜~ 10~ 𝐿.\)

\(\triangle 𝑉 = (10 – 2) 𝐿 = 8 𝐿\)

\(\therefore 𝑊=−1\times 8=− 8 ~𝑎𝑡𝑚 ~𝐿\)

Hence, \(−8 ~𝑎𝑡𝑚~ 𝐿\) work is done.

Step 2: Absorbed heat

We know from first law of thermodynamics,

\(\triangle U = q + W\)

As the system is working at constant temperature, i.e., isothermally.

\(\therefore \triangle U = 0\)

Hence, \(𝑞=- 𝑊=- (- 8~ 𝑎𝑡𝑚~ 𝐿)\)

Therefore, \(\rm8 ~atm ~L\) heat is absorbed in the expansion.

Final answer:

\(\rm ~-8~ atm~ L\) work is done, and \(\rm8~ atm~ L\) heat is absorbed in the expansion.