3
$\begingroup$

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?

$\endgroup$
7
  • 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

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.