A hot air ballon contains 270L of helium at 24C and 845 mmHg. WHat will be the volume of the ballon at -50c and pressure of 0.735atm?

Using the ideal gas law,

PV = nRT

where R = 0.08206 L•atm/(mol•°K), solving for n gives

n = PV/(RT)

n = (845 mmHg) (270 L) / ((0.08206 L•atm/(mol•°K)) (24 °C))

Convert the given temperature to °K and the given pressure to atm:

24 °C = (273.15 + 24) °K ≈ 297.2 °K

(845 mmHg) × (1/760 atm/mmHg) ≈ 1.11 atm

Then the balloon contains

n = (1.11 atm) (270 L) / ((0.08206 L•atm/(mol•°K)) (297.2 °K))

n ≈ 12.3 mol

of He.

Solve the same equation for V :

V = nRT/P

Convert the target temperature to °K:

-50 °C = (273.15 - 50) °K = 223.15 °K

Then the volume under the new set of conditions is

V = (12.3 mol) (0.08206 L•atm/(mol•°K)) (223.15 °K) / (0.735 atm)

V ≈ 306 L

b.

the accuracy of the data.

the answer is true. hope it .