Question

A balloon with a volume of 5.58 L is at a pressure of 0.86 atm and a temperature of 12oC. If the pressure is increased to 4.2 atm and the temperature is raised to 48oC, what is the new volume of the balloon?

Answer 1

A balloon with a volume of $5.58\mathrm{L}$ is at a pressure of 0.86 atm and a temperature of $12^{\circ}\mathrm{C}$. If the pressure is increased to 4.2 atm and the temperature is raised to $48^{\circ}\mathrm{C}$, what is the new volume of the balloon?

$$\begin{array}{l} V_{1} = 5.58 \mathrm{L} = 0.00558 \mathrm{m}^{3} \\ T_{1} = 12^{\circ} \mathrm{C} = 275 \mathrm{K} \\ P_{1} = 0.86 \mathrm{atm} = 87139.5 \mathrm{Pa} \\ T_{2} = 48^{\circ} \mathrm{C} = 321 \mathrm{K} \\ P_{2} = 4.2 \mathrm{atm} = 425565 \mathrm{Pa} \\ V_{2} - ? \\ \end{array}$$ $$\begin{array}{l} P_{1} V_{1} = n R T_{1} \\ P_{2} V_{2} = n R T_{2} \\ P_{1} V_{1} / P_{2} V_{2} = n R T_{1} / n R T_{2} \\ P_{1} V_{1} T_{2} = P_{2} V_{2} T_{1} \\ V_{2} = P_{1} V_{1} T_{2} / P_{2} T_{1} \\ V_{2} = (87139.5 \mathrm{Pa} \times 0.00558 \mathrm{m}^{3} \times 321 \mathrm{K}) / (425565 \mathrm{Pa} \times 275 \mathrm{K}) = 0.00133 \mathrm{m}^{3} \\ \end{array}$$

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