NTA Abhyas JEE Main2020PhysicsThermodynamicsPractice
The specific heats of an ideal gas at constant pressure and constant volume are 525 J kg ° C - 1 and 315 J kg ° C - 1 respectively. Its density at NTP is
Options
- A0.64 k g m - 3
- B1.20 k g m - 3
- C1.75 k g m - 3
- D2.62 k g m - 3
Correct answer
C. 1.75 k g m - 3
Step-by-step solution
If M is molecular mass of the gas, then from M C p - C v = R M = 8.31 210 = 0.0392 If ρ is density of the gas at NTP, then mass of 1 m 3 of gas at NTP = ρ kg ∴ Mass of 22.4 L ( = 22.4 × 10 - 3 m 3 ) of gas at NTP = ρ × 22.4 × 10 - 3 kg, which is the molecular mass of the gas ∴ ρ × 22.4 × 10 - 3 = 0.0392 ρ = 0.0392 22.4 × 10 - 3 = 1.75 k g m - 3