Question:
A bubble of an ideal gas rises from the bottom of a lake to the surface. At the bottom, the pressure is \(3\ \text{atm}\) and the temperature is \(7^\circ\text{C}\). At the surface, the pressure is \(1\ \text{atm}\) and the temperature is \(27^\circ\text{C}\). If the initial volume of the bubble was \(V_0\), what is its volume \(V\) at the surface?
A bubble of an ideal gas rises from the bottom of a lake to the surface. At the bottom, the pressure is \(3\ \text{atm}\) and the temperature is \(7^\circ\text{C}\). At the surface, the pressure is \(1\ \text{atm}\) and the temperature is \(27^\circ\text{C}\). If the initial volume of the bubble was \(V_0\), what is its volume \(V\) at the surface?
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In gas law questions, always convert Celsius temperature into kelvin before calculation.
Updated On: May 7, 2026
- \(3V_0\)
- \(3.21V_0\)
- \(0.9V_0\)
- \(5.4V_0\)
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The Correct Option is B
Solution and Explanation
Concept:
For a fixed amount of ideal gas:
\[
\frac{PV}{T}=\text{constant}
\]
Temperature must be converted into kelvin.
Step 1: Given initial conditions at the bottom: \[ P_1=3\ \text{atm} \] \[ T_1=7^\circ\text{C}=7+273=280\ \text{K} \] \[ V_1=V_0 \]
Step 2: Final conditions at the surface: \[ P_2=1\ \text{atm} \] \[ T_2=27^\circ\text{C}=27+273=300\ \text{K} \] Let final volume be \(V_2=V\).
Step 3: Use combined gas law. \[ \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} \] \[ V_2=\frac{P_1V_1T_2}{P_2T_1} \]
Step 4: Substitute values. \[ V=\frac{3\times V_0\times 300}{1\times 280} \] \[ V=\frac{900}{280}V_0 \] \[ V=3.21V_0 \] Therefore, \[ \boxed{3.21V_0} \]
Step 1: Given initial conditions at the bottom: \[ P_1=3\ \text{atm} \] \[ T_1=7^\circ\text{C}=7+273=280\ \text{K} \] \[ V_1=V_0 \]
Step 2: Final conditions at the surface: \[ P_2=1\ \text{atm} \] \[ T_2=27^\circ\text{C}=27+273=300\ \text{K} \] Let final volume be \(V_2=V\).
Step 3: Use combined gas law. \[ \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} \] \[ V_2=\frac{P_1V_1T_2}{P_2T_1} \]
Step 4: Substitute values. \[ V=\frac{3\times V_0\times 300}{1\times 280} \] \[ V=\frac{900}{280}V_0 \] \[ V=3.21V_0 \] Therefore, \[ \boxed{3.21V_0} \]
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