Question

A 12 cm diameter piston-cylinder device contains air at a pressure of 100 kPa at 24oC. The piston is initially 20 cm from the base of the cylinder. The gas is now compressed and 0.1 kJ of boundary work is added to the gas. The temperature of the gas remains constant during this process.
a. How much heat was transferred to/from the gas?
b. What is the final volume and pressure in the cylinder?
c. Find the change in entropy of the gas. Why is this value negative if entropy always increases in actual processes?

Answer 1

Answer:

ΔQ = 0.1 kJ

$\mathbf{v_f = 1.445*10^{-3} m^3}$

$\mathbf{P_f = 156.5 \ kPa}$

ΔS = -0.337 J/K

The value negative is due to the fact that there is need to be the same amount of positive change in surrounding as a result of compression.

Explanation:

GIven that:

Diameter of the piston-cylinder = 12 cm

Pressure of the piston-cylinder = 100 kPa

Temperature =24 °C

Length of the piston = 20 cm

Boundary work ΔW = 0.1 kJ

The gas is compressed and The temperature of the gas remains constant during this process.

We are to find ;

a. How much heat was transferred to/from the gas?

According to the first law of thermodynamics ;

ΔQ = ΔU + ΔW

Given that the temperature of the gas remains constant during this process; the isothermal process at this condition ΔU = 0.

Now

ΔQ = ΔU + ΔW

ΔQ = 0 + 0.1 kJ

ΔQ = 0.1 kJ

Thus; the amount of heat that was transferred to/from the gas is : 0.1 kJ

b. What is the final volume and pressure in the cylinder?

In an isothermal process;

Workdone W = $\int dW$

$W = \int pdV \\ \\ \\W = \int \dfrac{nRT}{V}dv \\ \\ \\ W = nRt \int \dfrac{dv}{V} \\ \\ \\ W = nRT In V |^{V_f} __{V_i}} \\ \\ \\ W = nRT \ In \dfrac{V_f}{V_i}$

Since the gas is compressed ; then $v_f< v_i$

However;

$W =- nRT \ In \dfrac{V_f}{V_i}$

$W =- P_1V_1 \ In \dfrac{V_f}{V_i}$

The initial volume for the cylinder is calculated as ;

$v_1 = \pi r^2 h \\ \\ v_1 = \pi r^2 L \\ \\ v_1 = 3.14*(6*10^{-2})^2*(20*10^{-2}) \\ \\ v_1 = 2.261*10^{-3} \ m^3$

Replacing over values into the above equation; we have :

$100 = - ( 100*10^3 *2.261*10^{_3}) In (\dfrac{v_f}{v_i}) \\ \\ - In (\dfrac{v_f}{v_i})= \dfrac{100}{(100*10^3*2.261*10^{-3})} \\ \\ - In \ v_f + In \ v_i = \dfrac{100}{226.1} \\ \\ - In \ v_f = - In \ v_i + \dfrac{100}{226.1} \\ \\ - In \ v_f = - In (2.261*10^{-3} + \dfrac{100}{226.1 } \\ \\ - In \ v_f = 6.1 + 0.44 \\ \\ - In \ v_f = 6.54 \\ \\ - In \ v_f = -6.54 \\ \\ v_f = e^{-6.54} \\ \\ \mathbf{v_f = 1.445*10^{-3} m^3}$

The final pressure can be calculated by using :

$P_1V_1 = P_2V_2 \\ \\ P_iV_i = P_fV_f$

$P_f =\dfrac{P_iV_i}{V_f}$

$P_f =\dfrac{100*2.261*10^{-3}}{1.445*10^{-3}}$

$P_f = 1.565*10^2 \ kPa$

$\mathbf{P_f = 156.5 \ kPa}$

c. Find the change in entropy of the gas. Why is this value negative if entropy always increases in actual processes?

The change in entropy of the gas is given  by the formula:

$\Delta S=\dfrac{\Delta Q}{T}$

where

T =  24 °C = (24+273)K

T = 297 K

$\Delta S=\dfrac{-100 \ J}{297 \ K}$

ΔS = -0.337 J/K

The value negative is due to the fact that there is need to be the same amount of positive change in surrounding as a result of compression.