정답: $4$

• $\mathrm{N}\mathrm{e}\backslash mathrm\{Ne\}$의 원자량은 $2020$이므로 (가)에서 $\mathrm{N}\mathrm{e}\backslash mathrm\{Ne\}$의 몰 수는 $2w\mathrm{g}\times {\displaystyle \frac{1}{20\mathrm{g}\mathrm{/}\mathrm{m}\mathrm{o}\mathrm{l}}}={\displaystyle \frac{w}{10}}\mathrm{m}\mathrm{o}\mathrm{l}2w\backslash ,\backslash mathrm\{g\}\; \backslash times\; \backslash dfrac\{1\}\{20\backslash ,\backslash mathrm\{g/mol\}\}\; =\; \backslash dfrac\{w\}\{10\}\backslash ,\backslash mathrm\{mol\}$이다.
• $\mathrm{A}\mathrm{r}\backslash mathrm\{Ar\}$의 원자량은 $4040$이므로 (나)에서 $\mathrm{A}\mathrm{r}\backslash mathrm\{Ar\}$의 몰 수는 $w\mathrm{g}\times {\displaystyle \frac{1}{40\mathrm{g}\mathrm{/}\mathrm{m}\mathrm{o}\mathrm{l}}}={\displaystyle \frac{w}{40}}\mathrm{m}\mathrm{o}\mathrm{l}w\backslash ,\backslash mathrm\{g\}\; \backslash times\; \backslash dfrac\{1\}\{40\backslash ,\backslash mathrm\{g/mol\}\}\; =\; \backslash dfrac\{w\}\{40\}\backslash ,\backslash mathrm\{mol\}$이다.

(가)와 (나)에서 각각 이상 기체 법칙을 적용하면 아래와 같다.

• (가)에서 이상 기체 법칙 적용:
  $P_{\text{(가)}}\times 1\mathrm{L}={\displaystyle \frac{w}{10}}\mathrm{m}\mathrm{o}\mathrm{l}\times R\times T\mathrm{K}P\_\{\backslash text\{(가)\}\}\; \backslash times\; 1\backslash ,\backslash mathrm\{L\}\; =\; \backslash dfrac\{w\}\{10\}\backslash ,\backslash mathrm\{mol\}\; \backslash times\; R\; \backslash times\; T\backslash ,\backslash mathrm\{K\}$,
  $P_{\text{(가)}}={\displaystyle \frac{w}{10}}RT\mathrm{m}\mathrm{o}\mathrm{l}\cdot \mathrm{K}\mathrm{/}\mathrm{L}P\_\{\backslash text\{(가)\}\}\; =\; \backslash dfrac\{w\}\{10\}RT\backslash ,\backslash mathrm\{mol·K/L\}$이다.

• (나)에서 이상 기체 법칙 적용:
  $P_{\text{(나)}}\times 3\mathrm{L}={\displaystyle \frac{w}{40}}\mathrm{m}\mathrm{o}\mathrm{l}\times R\times 2T\mathrm{K}P\_\{\backslash text\{(나)\}\}\; \backslash times\; 3\backslash ,\backslash mathrm\{L\}\; =\; \backslash dfrac\{w\}\{40\}\backslash ,\backslash mathrm\{mol\}\; \backslash times\; R\; \backslash times\; 2T\backslash ,\backslash mathrm\{K\}$,
  $P_{\text{(나)}}={\displaystyle \frac{w}{60}}RT\mathrm{m}\mathrm{o}\mathrm{l}\cdot \mathrm{K}\mathrm{/}\mathrm{L}P\_\{\backslash text\{(나)\}\}\; =\; \backslash dfrac\{w\}\{60\}RT\backslash ,\backslash mathrm\{mol·K/L\}$이다.

그러므로 ${\displaystyle \frac{P_{\text{(가)}}}{P_{\text{(나)}}}}=\backslash dfrac\{P\_\{\backslash text\{(가)\}\}\}\{P\_\{\backslash text\{(나)\}\}\}\; =\{\}$${\displaystyle \frac{{\displaystyle \frac{w}{10}}RT\mathrm{m}\mathrm{o}\mathrm{l}\cdot \mathrm{K}\mathrm{/}\mathrm{L}}{{\displaystyle \frac{w}{60}}RT\mathrm{m}\mathrm{o}\mathrm{l}\cdot \mathrm{K}\mathrm{/}\mathrm{L}}}=\backslash dfrac\{\backslash dfrac\{w\}\{10\}RT\backslash ,\backslash mathrm\{mol·K/L\}\}\{\backslash dfrac\{w\}\{60\}RT\backslash ,\backslash mathrm\{mol·K/L\}\}\; =\{\}$${\displaystyle \frac{w}{10}}RT\mathrm{m}\mathrm{o}\mathrm{l}\cdot \mathrm{K}\mathrm{/}\mathrm{L}\times {\displaystyle \frac{60}{wRT\mathrm{m}\mathrm{o}\mathrm{l}\cdot \mathrm{K}\mathrm{/}\mathrm{L}}}=\backslash dfrac\{w\}\{10\}RT\backslash ,\backslash mathrm\{mol·K/L\}\; \backslash times\; \backslash dfrac\{60\}\{wRT\backslash ,\backslash mathrm\{mol·K/L\}\}\; =\{\}$$66$이다.

따라서 답은 $44$번이다.

정답: $5$

$\mathrm{N}\mathrm{e}\backslash mathrm\{Ne\}$의 원자량은 $2020$이므로 실린더 속에 들어 있는 $\mathrm{N}\mathrm{e}\backslash mathrm\{Ne\}$의 몰 수는 $30\mathrm{g}\times {\displaystyle \frac{1}{20\mathrm{g}\mathrm{/}\mathrm{m}\mathrm{o}\mathrm{l}}}={\displaystyle \frac{3}{2}}\mathrm{m}\mathrm{o}\mathrm{l}30\backslash ,\backslash mathrm\{g\}\; \backslash times\; \backslash dfrac\{1\}\{20\backslash ,\backslash mathrm\{g/mol\}\}\; =\; \backslash dfrac\{3\}\{2\}\backslash ,\backslash mathrm\{mol\}$이다.

ㄱ. (가)에서 이상 기체 법칙을 적용하면
$1\mathrm{a}\mathrm{t}\mathrm{m}\times 3\mathrm{L}={\displaystyle \frac{3}{2}}\mathrm{m}\mathrm{o}\mathrm{l}\times R\times T_1\mathrm{K}1\backslash ,\backslash mathrm\{atm\}\; \backslash times\; 3\backslash ,\backslash mathrm\{L\}\; =\; \backslash dfrac\{3\}\{2\}\backslash ,\backslash mathrm\{mol\}\; \backslash times\; R\; \backslash times\; T\_\{1\}\backslash ,\backslash mathrm\{K\}$,
$R={\displaystyle \frac{2}{T_1}}\mathrm{a}\mathrm{t}\mathrm{m}\cdot \mathrm{L}\mathrm{/}(\mathrm{m}\mathrm{o}\mathrm{l}\cdot \mathrm{K})R\; =\; \backslash dfrac\{2\}\{T\_\{1\}\}\backslash ,\backslash mathrm\{atm·L/(mol·K)\}$이다. (O)

ㄴ. (나)에서 $\mathrm{N}\mathrm{e}(g)\backslash mathrm\{Ne\}(g)$의 압력을 $PP$라 할 때 이상 기체 법칙을 적용하면
$P\times 2\mathrm{L}={\displaystyle \frac{3}{2}}\mathrm{m}\mathrm{o}\mathrm{l}\times R\times T_1\mathrm{K}P\; \backslash times\; 2\backslash ,\backslash mathrm\{L\}\; =\; \backslash dfrac\{3\}\{2\}\backslash ,\backslash mathrm\{mol\}\; \backslash times\; R\; \backslash times\; T\_\{1\}\backslash ,\backslash mathrm\{K\}$,
$P={\displaystyle \frac{3}{4}}R{T}_1\mathrm{m}\mathrm{o}\mathrm{l}\cdot \mathrm{K}\mathrm{/}\mathrm{L}P\; =\; \backslash dfrac\{3\}\{4\}RT\_\{1\}\backslash ,\backslash mathrm\{mol·K/L\}$이다.

여기서 ㄱ에서 구한 $R={\displaystyle \frac{2\mathrm{a}\mathrm{t}\mathrm{m}\cdot \mathrm{L}}{T_1\mathrm{m}\mathrm{o}\mathrm{l}\cdot \mathrm{K}}}R\; =\; \backslash dfrac\{2\backslash ,\backslash mathrm\{atm·L\}\}\{T\_\{1\}\backslash ,\backslash mathrm\{mol·K\}\}$를 대입하면
$P={\displaystyle \frac{3}{4}}\times {\displaystyle \frac{2\mathrm{a}\mathrm{t}\mathrm{m}\cdot \mathrm{L}}{T_1\mathrm{m}\mathrm{o}\mathrm{l}\cdot \mathrm{K}}}\times T_1\mathrm{m}\mathrm{o}\mathrm{l}\cdot \mathrm{K}\mathrm{/}\mathrm{L}P\; =\; \backslash dfrac\{3\}\{4\}\; \backslash times\; \backslash dfrac\{2\backslash ,\backslash mathrm\{atm·L\}\}\{T\_\{1\}\backslash ,\backslash mathrm\{mol·K\}\}\; \backslash times\; T\_\{1\}\backslash ,\backslash mathrm\{mol·K/L\}$,
$P={\displaystyle \frac{3}{2}}\mathrm{a}\mathrm{t}\mathrm{m}P\; =\; \backslash dfrac\{3\}\{2\}\backslash ,\backslash mathrm\{atm\}$이다. (O)

ㄷ. (다)에서 이상 기체 법칙을 적용하면
${\displaystyle \frac{3}{2}}\mathrm{a}\mathrm{t}\mathrm{m}\times 3\mathrm{L}={\displaystyle \frac{3}{2}}\mathrm{m}\mathrm{o}\mathrm{l}\times R\times T_2\mathrm{K}\backslash dfrac\{3\}\{2\}\backslash ,\backslash mathrm\{atm\}\; \backslash times\; 3\backslash ,\backslash mathrm\{L\}\; =\; \backslash dfrac\{3\}\{2\}\backslash ,\backslash mathrm\{mol\}\; \backslash times\; R\; \backslash times\; T\_\{2\}\backslash ,\backslash mathrm\{K\}$,
$T_2={\displaystyle \frac{3\mathrm{a}\mathrm{t}\mathrm{m}\cdot \mathrm{L}}{R\mathrm{m}\mathrm{o}\mathrm{l}\cdot \mathrm{K}}}T\_\{2\}\; =\; \backslash dfrac\{3\backslash ,\backslash mathrm\{atm·L\}\}\{R\backslash ,\backslash mathrm\{mol·K\}\}$이다.
여기서 ㄱ에서 구한 $R={\displaystyle \frac{2\mathrm{a}\mathrm{t}\mathrm{m}\cdot \mathrm{L}}{T_1\mathrm{m}\mathrm{o}\mathrm{l}\cdot \mathrm{K}}}R\; =\; \backslash dfrac\{2\backslash ,\backslash mathrm\{atm·L\}\}\{T\_\{1\}\backslash ,\backslash mathrm\{mol·K\}\}$를 대입하면
$T_2={\displaystyle \frac{3\mathrm{a}\mathrm{t}\mathrm{m}\cdot \mathrm{L}}{\mathrm{m}\mathrm{o}\mathrm{l}\cdot \mathrm{K}}}\times {\displaystyle \frac{T_1\mathrm{m}\mathrm{o}\mathrm{l}\cdot \mathrm{K}}{2\mathrm{a}\mathrm{t}\mathrm{m}\cdot \mathrm{L}}}T\_\{2\}\; =\; \backslash dfrac\{3\backslash ,\backslash mathrm\{atm·L\}\}\{\backslash mathrm\{mol·K\}\}\; \backslash times\; \backslash dfrac\{T\_\{1\}\backslash ,\backslash mathrm\{mol·K\}\}\{2\backslash ,\backslash mathrm\{atm·L\}\}$,
$T_1={\displaystyle \frac{2}{3}}{T}_{2}T\_\{1\}\; =\; \backslash dfrac\{2\}\{3\}T\_\{2\}$이다. (O)

따라서 답은 $55$번이다.