This question requires the use of the universal gas law PV=nRT. Since we are given P, V, and T at the beginning, we can solve for n, the number of moles of helium gas in the balloon. Since the pressure is given in millimeters of mercury (mmHg), we first need...

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This question requires the use of the universal gas law PV=nRT. Since we are given P, V, and T at the beginning, we can solve for n, the number of moles of helium gas in the balloon. Since the pressure is given in millimeters of mercury (mmHg), we first need to convert that into atmospheres (remember 1 atm=760 mmHg):

775 mmHg * (1 atm/760 mmHg) = 1.02 atm

Now plug the different values into the gas equation:

PV=nRT

n = (PV)/(RT)

n = (1.02 atm * 6.65 L) / (0.0821 L*atm/mole*K * 296.15 K) = 0.279 moles

Since the number of moles of helium gas does not change as the balloon rises, we can re-insert this value back into the equation with the new temperature and pressure values to get the new volume of the balloon. Remember to convert mmHg into atmospheres.

613 mmHg * (1 atm/760 mmHg) = 0.807 atm

PV=nRT

V = (nRT)/P

V = (0.279 moles * 0.0821 L*atm/mole*K * 283.65 K) / 0.807 atm = 8.05 liters

**The volume of the balloon at the new altitude is 8.05 liters.**