Gas is kept in a 0.1 m diameter cylinder under the weight of a 100 kg piston that is held down by a spring with a stiffness k = 5 kN / m. If the gauge pressure of the gas is 300 kPa, how much is the spring compressed?

For equilibrium of the gas and the piston the pressure exerted by the gas on the piston should be equal to the sum of weight of the piston and the force the spring exerts on the piston

Mathematically we can write

[tex]Force_{pressure}=Force_{spring}+Weight_{piston}[/tex]

we know that

[tex]Force_{pressure}=Pressure\times Area=300\times 10^{3}\times \frac{\pi \times 0.1^2}{4}=750\pi Newtons[/tex]

[tex]Weight_{piston}=mass\times g=100\times 9.81=981Newtons[/tex]

Now the force exerted by an spring compressed by a distance 'x' is given by [tex]Force_{spring}=k\cdot x=5\times 10^{3}\times x[/tex]

Using the above quatities in the above relation we get

[tex]5\times 10^{3}\times x+981=750\pi \\\\\therefore x=\frac{750\pi -981}{5\times 10^{3}}=0.275meters[/tex]