
A boy blows a balloon of volume 1 liter to pressure of 10 atmospheres in 10 sec. What is the power used to fill the balloon?
$\text{A}\text{. 10}\text{.13 W}$
$\text{B}\text{. 1}\text{.013 W}$
$\text{C}\text{. 101}\text{.3 W}$
$\text{D}\text{. 0}\text{.506 W}$
Answer
569.4k+ views
Hint: When the boy fills the balloon, he does work to fill the air. The work done to increase the volume of the balloon is given by product of pressure and change in volume.
Power is the rate at which the work is done.
Formula Used:
Work done, $W=P\Delta V$; Power, $P=\dfrac{W}{t}$
Complete answer:
When the boy fills the balloon, he does work against force due to air in the balloon. This work done is given by product of pressure and change in the volume.
If $P$ is the pressure and $V$ is the volume of the balloon then, work done
$W=P\Delta V$
Therefore, work done in filling the balloon to 1 liter volume and 10 atmospheric pressure is
$W=P({{V}_{f}}-{{V}_{i}})$
Where ${{V}_{f}}$ and ${{V}_{i}}$ are the final and initial volume of the balloon.
Initially the balloon is not inflated, so ${{V}_{i}}=0L$
Boy blows the balloon to 1 liter, thus ${{V}_{f}}=1L$
Change in volume, $\Delta V=({{V}_{f}}-{{V}_{i}})=(1-0)L$
$\Rightarrow \Delta V=1L={{10}^{-3}}{{m}^{3}}$
Pressure, P = $10\times 1.013\times {{10}^{5}}=1.013\times {{10}^{6}}Pa$
Hence,
$W=1.013\times {{10}^{6}}\times {{10}^{-3}}J$
Or $W=1013J$
Power is defined as the rate of doing work. In other words, Power is the amount of energy transferred or absorbed per unit time. Mathematically, we can write power as
$P=\dfrac{W}{t}$
Where W is the work done and t is the time taken in doing the work.
Here, the boy took 10 seconds to fill the balloon. Therefore,
P = $\dfrac{1013}{10}J/s$
$P=101.3watt$
Hence, option C is correct.
Note:
According to kinetic theory of gas, gas particles exert pressure on the walls of the container in which the gas is filled. Therefore, blowing up a balloon requires force to add air particles from our lungs into the balloon. These particles strike on the inside walls of the balloon creating enough air pressure to force the rubber of the balloon to expand and the balloon to inflate.
Power is the rate at which the work is done.
Formula Used:
Work done, $W=P\Delta V$; Power, $P=\dfrac{W}{t}$
Complete answer:
When the boy fills the balloon, he does work against force due to air in the balloon. This work done is given by product of pressure and change in the volume.
If $P$ is the pressure and $V$ is the volume of the balloon then, work done
$W=P\Delta V$
Therefore, work done in filling the balloon to 1 liter volume and 10 atmospheric pressure is
$W=P({{V}_{f}}-{{V}_{i}})$
Where ${{V}_{f}}$ and ${{V}_{i}}$ are the final and initial volume of the balloon.
Initially the balloon is not inflated, so ${{V}_{i}}=0L$
Boy blows the balloon to 1 liter, thus ${{V}_{f}}=1L$
Change in volume, $\Delta V=({{V}_{f}}-{{V}_{i}})=(1-0)L$
$\Rightarrow \Delta V=1L={{10}^{-3}}{{m}^{3}}$
Pressure, P = $10\times 1.013\times {{10}^{5}}=1.013\times {{10}^{6}}Pa$
Hence,
$W=1.013\times {{10}^{6}}\times {{10}^{-3}}J$
Or $W=1013J$
Power is defined as the rate of doing work. In other words, Power is the amount of energy transferred or absorbed per unit time. Mathematically, we can write power as
$P=\dfrac{W}{t}$
Where W is the work done and t is the time taken in doing the work.
Here, the boy took 10 seconds to fill the balloon. Therefore,
P = $\dfrac{1013}{10}J/s$
$P=101.3watt$
Hence, option C is correct.
Note:
According to kinetic theory of gas, gas particles exert pressure on the walls of the container in which the gas is filled. Therefore, blowing up a balloon requires force to add air particles from our lungs into the balloon. These particles strike on the inside walls of the balloon creating enough air pressure to force the rubber of the balloon to expand and the balloon to inflate.
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