Question

The volume of 1.5 kg of helium in a frictionless piston-cylinder device is initially 6 m3. Now, helium is compressed to 2 m3 while its pressure is maintained constant at 200 kPa. Determine the initial and final temperatures of helium, as well as the work required to compress it, in kJ.

Answer 1

Answer:

The initial temperature will be "385.1°K" as well as final will be "128.3°K".

Explanation:

The given values are:

Helium's initial volume, v₁ = 6 m³

Mass, m = 1.5 kg

Final volume, v₂ = 2 m³

Pressure, P = 200 kPa

As we know,

Work, $W=p(v_{2}-v_{1})$

On putting the estimated values, we get

⇒            $=200000(2-6)$

⇒            $=200000\times (-4)$

⇒            $=800,000 \ N.m$

Now,

Gas ideal equation will be:

⇒  $pv_{1}=mRT_{1}$

On putting the values. we get

⇒  $200000\times 6=1.5\times 2077\times T_{1}$

⇒  $T_{1}=\frac{1200000}{3115.5}$

⇒       $=385.1^{\circ}K$ (Initial temperature of helium)

and,

⇒  $pv_{2}=mRT_{2}$

On putting the values, we get

⇒  $200000\times 2=1.5\times 2077\times T_{2}$

⇒  $T_{2}=\frac{400000}{3115.5}$

⇒       $=128.3^{\circ}K$ (Final temperature of helium)