Question

An insulated rigid tank is divided into two equal parts by a partition. Initially, one part contains 4 kg of an ideal gas at 700 kPa and 59°C, and the other part is evacuated. The partition is now removed, and the gas expands into the entire tank. Determine the final temperature and pressure in the tank.

Answer 1

Answer:

$P_2=350\ kPa$

$T_2=59^{\circ}\ C$

Explanation:

Given that

mass , m = 4 kg

Initial pressure ,$P_1=700\ kPa$

Initial temperature ,$T_1=59^{\circ}\ C$

The volume of rigid tanks are same

$V_1=V$

$V_2=2 V$

Let's take final temperature$=T_2$

Given that tank is insulated that is why heat transfer in the tank will be zero.

By using energy balance

$E_{in}-E_{out}=\Delta U$

$\Delta U$= Change in the internal energy of the gas

$0 = m C_V(T_2-T_1$)           ( Cv=Specific heat capacity at constant volume)

$0 = T_2-T_1$

Therefore $T_1=T_2$

$T_2=59^{\circ}\ C$

We know that ideal gas equation for gas

P V = m R T

P=pressure ,V=Volume ,m=mass ,R= gas constant ,T=temperature

By using mass conservation

$m=\dfrac{P_1V_1}{RT_1}=\dfrac{P_2V_2}{RT_2}$

Now by putting the values in the above equation

$\dfrac{700\times V}{RT_1}=\dfrac{P_2\times 2V}{RT_1}$

$P_2=\dfrac{700}{2}\ kPa$

$P_2=350\ kPa$

Therefore the final volume will be 350 kPa and temperature will be 59°C.