Question

A gas is placed in a storage tank at a pressure of 30.0 atm at 20.3 C. As a safety device, a small metal plug in the tank is made of a metal alloy that melts at 130 C. If the tank is heated, what is the maximum pressure (in atm) that will be attained in the tank before the plug melts and releases gas?

Answer 1

Answer: 41.2 atm

Explanation

To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=30.0atm\\T_1=20.3^0C=(20.3+273)=293.3K\\P_2=?\\T_2=130^0C=(130+273)K=403K$

Putting values in above equation, we get:

$\frac{30.0}{293.3K}=\frac{P_2}{403}\\\\P_2=41.2atm$

The maximum pressure (in atm) that will be attained in the tank before the plug melts and releases gas is 41.2

Answer 2

The maximum pressure (in atm) that will be attained in the tank before the plug melts and releases gas is 41.22 atm

From the question given above, the following data were obtained:

Initial pressure (P₁) = 30 atm

Initial temperature (T₁) = 20.3 °C = 20.3 + 273 = 293.3 K

Final temperature (T₂) = 130 °C = 130 + 273 = 403 K

Final pressure (P₂) =?

The final pressure can be obtained as illustrated below:

$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}\\\\\frac{30}{293.3} = \frac{P_{2}}{403}$

Cross multiply

293.3 × P₂ = 30 × 403

293.3 × P₂ = 12090

Divide both side by 293.3

$P_{2} = \frac{12090}{293.3}$

P₂ = 41.22 atm

Therefore, the maximum pressure (in atm) that will be attained in the tank before the plug melts and releases gas is 41.22 atm

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