NEET-XII-Physics
24: Kinetic Theory of Gases
- #7A gas cylinder has walls that can bear a maximum pressure of 1.0 × 106 Pa. It contains a gas at 8.0 × 105 Pa and 300 K. The cylinder is steadily heated. Neglecting any change in the volume, calculate the temperature at which the cylinder will break.Ans : Given:
Maximum pressure that the cylinder can bear, Pmax = 1.0 × 106 Pa
Pressure in the gas cylinder, P1 = 8.0 × 105 Pa
Temperature in the cylinder, T1 = 300 K
Let T2 be the temperature at which the cylinder will break.
Volume is constant. Thus, (Given)
V1= V2 = V
Applying the five variable gas equation, we get
`` \frac{{P}_{1}V}{{T}_{1}}=\frac{{P}_{2}V}{{T}_{2}}(\because {\,\mathrm{\,V\,}}_{1}={\,\mathrm{\,V\,}}_{2})``
`` \Rightarrow \frac{{P}_{1}}{{T}_{1}}=\frac{{P}_{2}}{{T}_{2}}``
`` \Rightarrow {T}_{2}=\frac{{P}_{2}\times {T}_{1}}{{P}_{1}}``
`` \Rightarrow {T}_{2}=\frac{1.0\times {10}^{6}\times 300}{8.0\times {10}^{5}}=375\,\mathrm{\,K\,}``
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