Question

Ratio of $C_p$ and [tex]C_v[/tex of a gas ‘X’ is 1.4. The number of atoms of the gas ‘X’ present in 11.2 litres of it at NTP will be
(a) 6.02 ×10²³
(b) 1.2 × 10²³
(c) 3.01 × 10²³
(d) 2.01 × 10²³

Answer 1

Here is your answer ⤵⤵⤵

(a) 6.02 ×10²³

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Answer 2

$<b>$
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Ratio of $C_p$ and [tex]C_v[/tex of a gas ‘X’ is 1.4. The number of atoms of the gas ‘X’ present in 11.2 litres of it at NTP will be
(a) 6.02 ×10²³
(b) 1.2 × 10²³
(c) 3.01 × 10²³
(d) 2.01 × 10²³

→ [B]


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