# Isothermal expansion of an ideal gas calculated two ways

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)

(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?

• 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.
– Karsten
Commented Jan 11 at 16:54
• Also, I'm surprised you have a correction factor of 2.303 in there, given that you are using the natural logarithm.
– Karsten
Commented Jan 11 at 16:55
• 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 Commented Jan 11 at 17:03
• 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 ? Commented Jan 11 at 17:26
• 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 !. Commented Jan 11 at 17:44