what volume should a 1.0L balloon at 25 degrees celcius and 1.00 atmosphere pressure be if immersed in liquid nitrogen (-196 degrees celsius, 1atm)?

Answered below.

To determine the volume of the balloon at a lower temperature and under a different pressure, we need to apply Charles' Law and the Ideal Gas Law.

1. Charles' Law states that for a given amount of gas at constant pressure, the volume is directly proportional to the temperature in Kelvin (K). The formula is:

V1 / T1 = V2 / T2

Where:
V1 = initial volume of the balloon (1.0 L)
T1 = initial temperature of the balloon in Kelvin (25°C + 273.15)
V2 = final volume of the balloon (unknown)
T2 = final temperature of the balloon in Kelvin (-196°C + 273.15)

2. The Ideal Gas Law relates the pressure (P), volume (V), temperature (T), and the number of moles of gas (n). The formula is:

PV = nRT

Where:
P = pressure
V = volume
n = number of moles of gas
R = ideal gas constant
T = temperature in Kelvin

3. Since we are dealing with the same amount of gas (assuming air inside the balloon), the number of moles (n) and the ideal gas constant (R) will remain constant. Therefore, we can rewrite the Ideal Gas Law as:

P1V1 / T1 = P2V2 / T2

Where:
P1 = initial pressure in atmospheres (1.00 atm)
P2 = final pressure in atmospheres (1.00 atm)
V1 = initial volume of the balloon (1.0 L)
T1 = initial temperature of the balloon in Kelvin (25°C + 273.15)
V2 = final volume of the balloon (unknown)
T2 = final temperature of the balloon in Kelvin (-196°C + 273.15)

Now, we can calculate the unknown value, which is the final volume (V2).

Substituting the given values into the equation:

(1.00 atm)(1.0 L) / (25°C + 273.15) = (1.00 atm)(V2) / (-196°C + 273.15)

Simplifying the equation gives:

1.00 L / 298.15 K = V2 / 77.15 K

Cross-multiplying:

1.00 L * 77.15 K = V2 * 298.15 K

V2 = (1.00 L * 77.15 K) / 298.15 K

V2 ≈ 0.257 L

Therefore, the volume of the 1.0 L balloon, when immersed in liquid nitrogen at -196°C and 1.00 atm, would be approximately 0.257 liters.