Question

A piston/cylinder contains 1.5 kg of water at 200 kpa, 150 degrees celcius. It is now heated by a process in which pressure is linearly related to volume to a state of 600 kpa, 350 degrees celcius. Find the final volume and the work in the process.

Answer 1

Answer:  The final volume will be 719 L.

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at NTP, weighs equal to the molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

According to the ideal gas equation:

$PV=nRT$

P = Pressure of the gas = 200 kPa

V= Volume of the gas = ?

T= Temperature of the gas = 150°C = $(150+273)K= 423K$

R= Gas constant = 8.314 kPaL/K mol

n=  moles of gas= $\frac{1.5\times 1000g}{18g/mol}=83.3moles$

$V=\frac{nRT}{P}=\frac{83.3\times 8.314\times 423}{200}=1465L$

The combined gas equation is,:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 200 kPa

$P_2$ = final pressure of gas = 600 kPa

$V_1$ = initial volume of gas = 1465 L

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $150^oC=273+150=423K$

$T_2$ = final temperature of gas = $350^oC=273+350=623K$

Now put all the given values in the above equation, we get the final volume of gas.

$\frac{200\times 1465}{423K}=\frac{600\times V_2}{623K}$

$V_2=719L$

Therefore, the final volume will be 719 L.