Consider the following problem:

$\pu{1.0 mol}$ of a monoatomic ideal gas is expanded from state 1 to state 2 as shown in the figure. The magnitude of the work done for the expansion of gas from state 1 to state 2 at $\pu{300 K}$ is ..... $\pu{J}$. (Nearest integer)

State diagram showing curve from State 1 (6.0 pascals, 22.7 litres) down to State 2 (3 pascals)

(Given: $R=8.3 \,\mathrm {J\,K^{-1}\,mol^{-1}}$, $\ln 10 = 2.3$, $\log 2 = 0.30$)

I tried to solve this question with 2 methods but my answers were different.

  1. I used the equation for isothermal processes $$ w = 2.303nRT\log(v_2/v_1) \tag{1} $$ Considering $v_2$, $p_2$ (final volume and pressure) and $v_1$, $p_1$ as initial. Solving this way I got $\pu{1718.1 J}$.

  2. It's simply that I substituted $nRT$ in the above equation as $p_1V_1$ as $p_1V_1 = p_2V_2= nRT$ at constant temperature (isothermal). My answer came to be around $\pu{9397 J}$.

I'm not sure what I did wrong. Is my concept wrong?

  • 1
    $\begingroup$ Show us the steps. If the term $\ln \frac{V_2}{V_1}$ was used in both, and the numerical value of the factor you are multiplying with is the same, you should get the same answer. $\endgroup$
    – Karsten
    Jan 11 at 16:54
  • $\begingroup$ Also, I'm surprised you have a correction factor of 2.303 in there, given that you are using the natural logarithm. $\endgroup$
    – Karsten
    Jan 11 at 16:55
  • $\begingroup$ I mistyped I'm sorry. It's actually w=2.303nRTlog(V2/V1). The steps: since process is isothermal so T constt and work= 2.3×1×8.31×300 ×0.3 = 1720J (approx). Method 2 : P1=6 bar v1=22.7 so W=p1v1×2.3×log2=93.97 bar-L. To convert to J ×100 so 9397 J $\endgroup$
    – Minduelle
    Jan 11 at 17:03
  • 3
    $\begingroup$ Four questions. First, the expansion looks isothermal. Is it ? Nobody mentions it. Second, the curve should be a hyperbola. It is not. Why ? Third, P1 = 2·P2. Can we derive that V2 = 2·V1 ? Fourth, how can the volume of 1 mole gas be 22.7 L at 300 K and 6 bar ? $\endgroup$
    – Maurice
    Jan 11 at 17:26
  • 5
    $\begingroup$ The given data are wrong. I explain. The state 1 is : n = 1 mol, P = 6 bar, T = 300K, V = 22.7 L. With these values, the product pV (= 6 10^5 ·0.0227 = 13620 J ) is not equal to nRT (= 8.314 300 = 2494 J). pV is 6 times too high !. $\endgroup$
    – Maurice
    Jan 11 at 17:44


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