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Suggest the pressure and temperature at which $1.0\ \mathrm{mol}$ of $\ce{H2S}$ will be in same state corresponding to $1.0\ \mathrm{mol}$ $\ce{N2}$ at $1\ \mathrm{atm}$ and $298\ \mathrm{K}$.


$$v_{\mathrm{rms}, \ce{N2}} = \sqrt{3RT\over M} = \sqrt{3\times 0.082 \times 298\over 28} = 2.61$$ $$V = {RT\over p} = 24.43\tag{1}$$

$$v_{\mathrm{rms}, \ce{H2S}} = \sqrt{3RT\over M} = \sqrt{3 \times 0.082 \times T\over 34} = \sqrt{0.0072\times T}$$

I equated $v_{\mathrm{rms}, \ce{H2S}}$ and $v_{\mathrm{rms}, \ce{N2}}$, just to get the temperature. For which I got $T = 946.125\ \mathrm{K}\tag{2}$.

I then used ideal gas equation to find pressure of $\ce{H2S}$ using the $(1),(2)$

$$p = {RT\over V} = {946.125 \times 0.082 \over 24.43 } = 3.175\ \mathrm{atm}$$


I calculated rms because I thought that two gases at with same velocity and volume will have same state, But frankly I did not understand what same state mean in the question and what I have to do in this question.

Answer in the book, $ p = 2.6\ \mathrm{atm}$, $T = 881\ \mathrm K$


Somebody please tell me as to what same state mean in the context of this question ?

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