Answer to Question #344858 in General Chemistry for Lona

Question

What volume did a helium-filled balloon have at 26.8 °C and 1.87 atm if its new volume was 63.2 mL at 2.22 atm and 9.6°C? O a. 79.6 O b. 209 O c. 70.7 O d. 56.5

Answer 1

In an isolated system, the mole number remains the same:

$n_1=n_2$

In the ideal gas law, the mole number is related to the temperature T, volume V, pressure P and gas constant R:

$pV=nRT$

$n=\frac{pV}{RT}$

Thus:

$\frac{p_1V_1}{RT_1} = \frac{p_2V_2}{RT_2}$

The initial volume V₁ can be deduced from the last equation:

$V_1 = \frac{p_2V_2T_1}{T_2p_1} = \frac{2.22 \times 63.2 \times 26.8}{9.6 \times 1.87} = 209 ml$

Answer: b 209 ml