MCQ
Ratio of ${C_p}$ and ${C_v}$ of a gas $X$ is $1.4$, the number of atom of the gas $‘X’$ present in $11.2$ litres of it at $NTP$ will be
- ✓$6.02 \times {10^{23}}$
- B$1.2 \times {10^{23}}$
- C$3.01 \times {10^{23}}$
- D$2.01 \times {10^{23}}$
✓
Answer
Correct option: A.
$6.02 \times {10^{23}}$
(a) $\frac{{{C_P}}}{{{C_V}}} = 1.4$ so, given gas is diatomic
$11.2\,L$ $ = 3.01 \times {10^{23}}$ molecules
$\therefore $ No. of atoms $ = 3.01 \times {10^{23}} \times 2$ $ = 6.023 \times {10^{23}}$ atoms
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