In a thermonuclear device, the pressure of 0.050 liters of gas within the bomb casing reaches 4 0 x 10^6 atm. When the bomb casing is destroyed by the explosion, the gas is released into the atmosphere where it reaches a pressure of 1,00 atm. What is the volume of the gas after the explosion?
0.05 * 4.0 * 10^6 = 2 * 10^5 = 200,000 Liters
To solve this problem, we can use Boyle's Law, which states that the product of pressure and volume is constant for a given amount of gas at a constant temperature.
According to Boyle's Law, we have:
(P1)(V1) = (P2)(V2)
Where:
P1 = initial pressure = 4.0 x 10^6 atm
V1 = initial volume = 0.050 liters
P2 = final pressure = 1.00 atm
V2 = final volume (what we need to find)
Plugging in the given values:
(4.0 x 10^6)(0.050) = (1.00)(V2)
Simplifying the equation:
V2 = (4.0 x 10^6)(0.050) / 1.00
V2 = 200,000 liters
Therefore, the volume of the gas after the explosion is 200,000 liters.
To find the volume of the gas after the explosion, you can use the ideal gas law. The ideal gas law is given by the equation:
PV = nRT
Where:
P = pressure of the gas
V = volume of the gas
n = number of moles of the gas
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature of the gas in Kelvin
In this case, we know the pressure and the volume of the gas both before and after the explosion. We can assume that the number of moles of the gas remains constant, as no information is given to suggest otherwise. The temperature also remains constant as it is not mentioned in the question.
Let's denote the volume of the gas before the explosion as V1 and the pressure as P1. Similarly, the volume after the explosion will be denoted as V2 and the pressure as P2.
Using the ideal gas law, we can write two equations:
P1 * V1 = nRT
P2 * V2 = nRT
Since n, R, and T are constant, we can write:
P1 * V1 = P2 * V2
Now we can plug in the given values:
P1 = 4.0 x 10^6 atm
V1 = 0.050 liters
P2 = 1.00 atm
Substituting these values into the equation, we get:
(4.0 x 10^6 atm) * (0.050 liters) = (1.00 atm) * V2
Simplifying:
V2 = (4.0 x 10^6 atm * 0.050 liters) / 1.00 atm
V2 = 200,000 liters
Therefore, the volume of the gas after the explosion is 200,000 liters.