A hot air balloon contains 65 L of air at ground level, where the conditions are 98 kPa and 27°C.
What will be the new volume of the balloon when it rises to an altitude where the conditions are 10 kPa and –10°C?
- 400 L
- 484 L
- 558 L
- 610 L
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A hot air balloon contains 65 L of air at ground level, where the conditions are 98 kPa and 27°C.
What will be the new volume of the balloon when it rises to an altitude where the conditions are 10 kPa and –10°C?
\(C\)
→ Using the Combined Gas Law: \(\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\)
→ \(T_1 = 300\ \text{K}\) and \(T_2 = 263\ \text{K}\)
→ \(V_2=\dfrac{P_1V_1T_2}{T_1P_2}=\dfrac{98 \times 65 \times 263}{300 \times 10}=558\ \text{L}\)
\(\Rightarrow C\)
A sample of nitrogen gas at a pressure of 400 kPa and a temperature of 60.0°C occupies a volume of 35.0 L.
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a. \(104.4\ \text{L}\)
b. \(5.06\ \text{mol}\)
c. \(141.8\ \text{g}\)
a. Using the Combined Gas Law: \(\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\)
\(T_1 = 60^{\circ}\text{C}=333\ \text{K}\)
\(V_2=\dfrac{P_1V_1T_2}{T_1P_2}=\dfrac{400 \times 35 \times 298}{333 \times 120}=104.4\ \text{L}\)
b. Using the Ideal Gas Law: \(PV=nRT\)
\(n=\dfrac{PV}{RT}=\dfrac{400 \times 35}{8.314 \times 333}=5.06\ \text{mol}\)
c. Nitrogen gas \(\ce{(N2)}\) has a molar mass of \(28.02\ \text{g mol}^{-1}\)
\(\ce{m(N2)}= n \times MM = 5.06 \times 28.02 = 141.8\ \text{g}\)
A sample of gas is stored in a container at a pressure of 4.00 kPa, a temperature of 280 K, and a volume of 8.00 L. If the temperature increases to 320 K and the pressure changes to 4.50 kPa, what will the volume be?
\(B\)
→ Using the Combined Gas Law: \(\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\)
\(V_2\) | \(=\dfrac{P_1 \times V_1 \times T_2}{T_1 \times P_2}\) | |
\(=\dfrac{4.00 \times 8 \times 320}{280 \times 4.5}\) | ||
\(=8.1\ \text{L}\) |
\(\Rightarrow B\)