A meteorologist balloon contains 250.0 L of He at 22.0 C and 740.0 mmHg. If the volume of the balloon can vary according to external conditions what volume would it occupy at an altitude at which the temperature is -52.0 C and the pressure is .750 atm?

A. 240 L
B. 145 L
C. 760 L
D. 184790 L

Option A) 240 L

Explanation:

Given that:

Initial Volume of helium (V1) = 250.0L

Initial Temperature of helium (T1) = 22.0°C

[Convert temperature in celsius to Kelvin by adding 273

(22.0°C + 273 = 295K)]

Initial Pressure of helium (P1) = 740 mmHg

[convert pressure in mmHg to atmosphere

If 760 mmHg = 1 atm

740 mmHg = 740/760 = 0.97 atm]

Final Volume of helium (V2) = ?

Final temperature T2 = -52.0°C

[Convert -52°C to Kelvin by adding 273

-52°C + 273 = 221K]

Final pressure of helium = 0.750 atm

Since pressure, volume and temperature are given, apply the combined gas equation

(P1V1)/T1 = (P2V2)/T2

(0.97 x 250.0)/295= (0.750 x V2)/221

242.5/295 = 0.750V2/221

Cross multiply

242.5 x 221 = 0.75V2 x 295

53592.5 = 221.25V2

V2 = 53592.5 / 221.25

V2 = 242.2 L (Rounded to the nearest tens, V2 becomes 240 L)

Thus, the new volume of helium at the altitude is 240 liters