On earth a tank is 10.0L in size, has a pressure of 50ATM and a temperature of 23 degrees celsius. When the rocket gets near the sun, the temperature increases to 225 degrees celsius. If the tank bursts at 100ATM, will it burst?

Question

On earth a tank is 10.0L in size, has a pressure of 50ATM and a temperature of 23 degrees celsius.  When the rocket gets near the sun, the temperature increases to 225 degrees celsius.  If the tank bursts at 100ATM, will it burst?

Wyatt Atkins · Accepted Answer

No, it will not burst.

The trick here is to recognize the fact that the amount of gas in the tank remains constant.

Moreover, the volume of the gas tank remains constant as well, which is why you're trying to determine whether or not it will burst at the new temperature.

So, when the amount of gas and the volume it occupies remain constant, pressure and temperature have a direct relationship - this is known as Gay Lussac's Law.

In other words, if temperature increases, then you can expect pressure to increase as well. Likewise, if temperature decreases, pressure will decrease as well.

This happens because as the temperature increases, the average kinetic energy of the gas molecules increases as well. As a result, the gas molecules will hit the walls of the container more frequently and with more force #-># pressure increases.

Mathematically, this is written as

#P_1/T_1 = P_2/T_2" "#, where
#P_1#, #T_1# - the pressure and temperature of the gas at an initial state;
#P_2#, #V_2# - the pressure and temperature of the gas at a final state.

So, plug in your values and solve for #P_2#, the pressure of the gas after the temperature is increased - do not forget that the temperatures of the gas must be expressed in Kelvin!

#P_1/T_1 = P_2/T_2 implies P_2 = T_2/T_1 * P_1#
#P_2 = ((273.15 + 225)color(red)(cancel(color(black)("K"))))/((273.15 + 23)color(red)(cancel(color(black)("K")))) * "50 atm" ~~ "84 atm"#

Since this pressure is smaller than the limit pressure of #"100 atm"#, it follows that the tank will not burst at #225^@"C"#.

Luna Chen · Answer

undefined To determine if the tank will burst, we need to calculate the final pressure of the gas inside the tank when the temperature increases from 23°C to 225°C. We can use the combined gas law, which states:

$$ \frac{P_1 \cdot V_1}{T_1} = \frac{P_2 \cdot V_2}{T_2} $$

Where:
- $ P_1 $ and $ P_2 $ are the initial and final pressures respectively,
- $ V_1 $ and $ V_2 $ are the initial and final volumes respectively,
- $ T_1 $ and $ T_2 $ are the initial and final temperatures respectively.

Given:
- Initial pressure, $ P_1 $ = 50 atm
- Initial volume, $ V_1 $ = 10.0 L
- Initial temperature, $ T_1 $ = 23°C = 23 + 273.15 = 296.15 K
- Final temperature, $ T_2 $ = 225°C = 225 + 273.15 = 498.15 K
- Bursting pressure, $ P_{\text{burst}} $ = 100 atm

We can rearrange the formula to solve for the final pressure $ P_2 $:

$$ P_2 = \frac{P_1 \cdot V_1 \cdot T_2}{V_2 \cdot T_1} $$

Since we are not given the final volume, we assume the volume remains constant (10.0 L):

$$ P_2 = \frac{50 \text{ atm} \cdot 10.0 \text{ L} \cdot 498.15 \text{ K}}{10.0 \text{ L} \cdot 296.15 \text{ K}} $$

$$ P_2 = \frac{249075 \text{ atm·L·K}}{2961.5 \text{ L·K}} $$

$$ P_2 = 84.06 \text{ atm} $$

The final pressure inside the tank when it gets near the sun would be approximately 84.06 atm, which is below the bursting pressure of 100 atm. Therefore, the tank will not burst.