Russell bradley carried 207 kg of bricks 3.65 m up a ladder. If the amount of work required to perform that task is used to compress a gas at a constant pressure of 1.8 X 10^6 Pa, what is the change in volume of the gas?

Respuesta :

Answer:

[tex]dV=4.1136\times 10^{-4}\ m^3[/tex]

Explanation:

Given:

  • Mass of the bricks carried, [tex]m=207\ kg[/tex]
  • height of displacement, [tex]h=3.65\ m[/tex]
  • Constant pressure of the gas, [tex]P=1.8\times 10^6\ Pa[/tex]

Now the work done to displace the brick along the length of the ladder:

[tex]W=(m.g)\times h[/tex]

[tex]W=(207\times 9.8)\times 3.65[/tex]

[tex]W=740.439\ J[/tex]

As we know that the work done in compressing the gas at constant pressure is given as:

[tex]W=P.dV[/tex]

where:

[tex]dV=[/tex] change in volume of the gas

[tex]740.439=1.8\times 10^6\times dV[/tex]

[tex]dV=4.1136\times 10^{-4}\ m^3[/tex]

The change in volume of the gas is [tex]4.1177\times10^{-3}\ m^3[/tex].

Given to us:

Mass of the brick m= 207 kg,

Height at which brick is displaced h= 3.65 m,

constant pressure p = [tex]1.8 X 10^6[/tex] Pa,

Acceleration due to gravity g=9.81 m/s²

To find out the work done by Russell to displace the brick,

[tex]\rm work\ done, w= (mgh)[/tex]

putting the numerical values we get,

[tex]\begin{aligned}w&=mgh\\&=207\times9.81\times3.65\\&= 7411.9455\ \rm J\\\end{aligned}[/tex]

For a constant pressure process work done is given by,

[tex]\begin{aligned}w&=p\delta v\\7411.9455&=1.8\times 10^6\times\delta v\\\delta v&=0.0041177\ m^3\\\delta v&=4.1177\times10^{-3}\ m^3\end{aligned}[/tex]

Hence, the change in volume of the gas is [tex]4.1177\times10^{-3}\ m^3[/tex].

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