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A sealed balloon is filled with 2.20 liters of helium gas at 20.0°

Cif the pressure does not change, what is the temperature when the balloon expands to a volume of 2.60 liters

Verified Answer :

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To solve this we assume
that the gas inside the balloon is an ideal gas. Then, we can use the ideal gas
equation which is expressed as PV = nRT. At a constant pressure and number of
moles of the gas the ratio T/V is equal to some constant. At another set of
condition of temperature, the constant is still the same. Calculations are as
follows:

T1 / V1 = T2 / V2

T2 = T1 x
V2 / V1 

T2 = 20 x 2.60 / 2.20

T2
= 23.64 degrees Celsius

Answer:

The temperature when the balloon expands to 2.60 L is 346 K

Explanation:

Given:

Initial volume of the ballon, V1 = 2.20 L

Initial Temperature, T1 = 20 C

Final volume of the ballon, V2 = 2.60 L

To determine:

Final Temperature, T2

Explanation:

Based on the ideal gas equation

\(PV = nRT\)

where P = pressure, V = volume ; n = moles of gas

R = gas constant, T = temperature

At constant n and P, the above equation becomes:

V/T = constant

This is the Charles law

Therefore:

\(\frac{V1}{T1} = \frac{V2}{T2} \\\\T2 = \frac{V2}{V1}*T1 = \frac{2.60L}{2.20L}*(20+273)K = 346 K\)

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