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)?
V1/T1 = V2/T2
Of course the liquid nitrogen would freeze the rubber balloon, it would become brittle and crack BUT I know this is just a problem in Charles' Law.
To find the volume of the balloon when immersed in liquid nitrogen, we can use the combined gas law formula:
P1 * V1 / T1 = P2 * V2 / T2
Where:
P1 = initial pressure (1.00 atm)
V1 = initial volume (1.0 L)
T1 = initial temperature (25°C + 273.15 = 298.15 K)
P2 = final pressure (1 atm)
T2 = final temperature (-196°C + 273.15 = 77.15 K)
Let's solve for V2:
(1.00 atm * 1.0 L) / 298.15 K = (1 atm * V2) / 77.15 K
(1.00 L * 77.15 K) / 298.15 K = V2
0.2588 L = V2
Therefore, the volume of the balloon when immersed in liquid nitrogen will be approximately 0.2588 liters.
To determine the volume of the balloon when immersed in liquid nitrogen, we can use the combined gas law, which relates the initial and final conditions of temperature and pressure to the initial and final volumes of the gas.
The combined gas law formula is as follows:
(P1 * V1) / (T1) = (P2 * V2) / (T2),
where:
P1 and P2 are the initial and final pressures,
V1 and V2 are the initial and final volumes,
T1 and T2 are the initial and final temperatures.
Let's assign the following values to the variables:
P1 = 1.00 atm,
V1 = 1.0 L,
T1 = 25 °C = 298 K,
P2 = 1.00 atm,
T2 = -196 °C = 77 K.
Now we can calculate the final volume (V2) of the balloon:
(V1) / (T1) = (P2 * V2) / (T2).
Substituting the given values:
(1.0 L) / (298 K) = (1.00 atm * V2) / (77 K).
Now, we can rearrange the equation to solve for V2:
V2 = (1.0 L * 1.00 atm * 77 K) / (298 K).
Calculating this expression:
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 pressure, would be approximately 0.257 L.