Question

A 24-cm-diameter vertical cylinder is sealed at the top by a frictionless 15 kg piston. The piston is 90 cm above the bottom when the gas temperature is 315 ∘C. The air above the piston is at 1.00 atm pressure.
A) What is the gas pressure inside the cylinder?B) What will the height of the piston be if the temperature is lowered to18 ∘C?

Answer 1

Answer:

(A) $P_i=3249.41\ Pa$

(B) $V_f=0.3358\ m$

Explanation:

Given:

• diameter of the cylinder, $d= 0.24\ m$

• mass of piston sealing on the top, $m_p=15\ kg$

• initial temperature of the piston, $T_i=315+273= 588\ K$

• initial height of piston, $h_i=0.9\ m$

• atmospheric pressure on the piston, $p_a=1 atm=101325\ Pa$

(A)

Initial pressure of gas is the pressure balanced by the weight of piston:

$P_i=\frac{m_p.g}{\pi.d^2\div 4}$

$P_i=\frac{15\times 9.8}{\pi\times (0.24^2\div 4)}$

$P_i=3249.41\ Pa$

Which is gauge pressure because it is measured with respect to the atmospheric pressure.

(B)

Given:

• Final temperature, $T_f=18+273=291\ K$

Now, volume of air initially in the cylinder:

$V_i=\pi.d.h_i$

$V_i=\pi\times 0.24\times 0.9$

$V_i=0.6786\ m^3$

Using gas law:

$\frac{P_i V_i}{T_i}= \frac{P_f V_f}{T_f}$ ........................................(1)

∵In every condition of equilibrium the gas pressure will be balanced by the weight of the piston so it is an isobaric transition.

∴$P_i=P_f$

Hence eq. (1) is reduced to:

$\frac{V_i}{T_i}= \frac{V_f}{T_f}$

putting respective values:

$\frac{0.6786}{588}= \frac{V_f}{291}$

$V_f=0.3358\ m$