Question

A 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.

Answer 1

The temperature at which the cylinder will break is about 375°K

Explanation:

Step 1:

Given data in the question  :

Highest pressure the cylinder can bear, Pmax = $1.0 \times 10^{5} \mathrm{Pa}$

the gas cylinder pressure, $P_{1}=8.0 \times 10^{8} \mathrm{Pa}$

cylinder temperature, $T_1$= 300 K

Let $T_2$ be the temperature a cylinder breaks at constant volume  

Step 2:

Thus ,                  

( Given )$V_1 = V_2 = V$

We are able to apply the five variable gas equation

$\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}$

Step 3:

On substituting the values ,We get  

$\frac{8 \times 10^{5} \times V}{300}=\frac{1 \times 10^{6} \times V}{T_{2}}$

$\frac{8 \times 10^{5}}{300}=\frac{10^{6}}{T_{2}}$        

$T_{2}=\frac{300 \times 10^{6}}{8 \times 10^{5}}$

$T_{2}=\frac{300 \times 10}{8}$

$T_{2}=\frac{3000}{8}$

$T_{2}=375^{\circ} K$

Answer 2

The temperature at which the cylinder will break is 300 K

Explanation:

Given data in the question  

Highest pressure the cylinder can bear, $\mathrm{P}_{\max }=1.0 \times 10^{6} \mathrm{Pa}$

the gas cylinder pressure, $P_{1}=8.0 \times 10^{5} \mathrm{Pa}$

cylinder temperature, $T_1 = 300 K$

Let be $T_2$ the temperature a cylinder breaks at constant volume  

Thus ,                  

( Given ) $V_1 = V_2 = V$

We are able to apply the five variable gas equation

$\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}$

On substituting the values ,We get  

$\frac{8 \times 10^{5} \times V}{300}=\frac{1 \times 10^{6} \times V}{T_{2}}$

$\frac{8 \times 10^{5}}{300}=\frac{10^{6}}{T_{2}}$

$T_{2}=\frac{300 \times 10^{6}}{8 \times 10^{5}}$

$T_{2}=\frac{3000}{8}$

$T_{2}=375 K$

The temperature at which the cylinder will break is 300 K which bears the maximum pressure of about 1.0 × 106 Pa.