Question: Ratio of Cₚ and Cᵥ 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 [CBSE 1999]
(a) 6.02 x 10²³
(b) 1.2 x 10²³
(c) 3.01 x 10²³
(d) 2.01 x 10²³
Correct Answer: (a) 6.02 x 10²³
![Question: Ratio of Cₚ and Cᵥ 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 [CBSE 1999] (a) 6.02 x 10²³ (b) 1.2 x 10²³ (c) 3.01 x 10²³ (d) 2.01 x 10²³](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhAZYV8C-11DXcMq5VwbYixTmYz8kOovoE3jkfhvm9aTPXwGTLMd0UcUtNWNZhVTXzEP1d8uvMtZy-NdlEezNH2PkCD9ugqXI1_C7dA6tPyFJKDoNuoB5l1QjkhqI9kz8FNPe7BJTtcvsOfP37GTrOh4K-0d9c0YexKDMkJp2igk8hDEgmMTT73o3VN/s1920/Ratio%20of%20C%E2%82%9A%20and%20C%E1%B5%A5%20of%20a%20gas%20X%20is%201.4,%20the%20number%20of%20atom%20of%20the%20gas%20%E2%80%98X%E2%80%99%20present%20in%2011.2%20litres%20of%20it%20at%20NTP%20will%20be%20%5BCBSE%201999%5D.jpg "Question: Ratio of Cₚ and Cᵥ 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 [CBSE 1999] (a) 6.02 x 10²³ (b) 1.2 x 10²³ (c) 3.01 x 10²³ (d) 2.01 x 10²³")
Ratio of Cₚ and Cᵥ 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 [CBSE 1999]