-
A diatomic gas initially at 18°C is compressed adiabatically to one eighth of its original volume. The temperature after compression will be
-
- 18°C
- 668.4°K
- 395.4°C
- 144°C
Correct Option: B
Initial temperature (T1) = 18°C = 291 K
Let Initial volume (V1) = V
| Final volume (V2) = | ||
| 8 |
According to adiabatic process, TVγ - 1 = constant
According to question, T1V1γ - 1 = T2V2γ - 1
| ⇒ T2 = 293 |  |  | γ - 1 | ||
| V2 |
⇒ T2 = 293(8)(7 / 5) - 1 = 293 × 2.297 = 668.4K
| For diatomic gas , γ = | = | | |||
| Cv | 5 |
