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

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

Step-by-step 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* (-4)`

⇒
`=800,000 \ N.m`

Now,

Gas ideal equation will be:

⇒
`pv_(1)=mRT_(1)`

On putting the values. we get

⇒
`200000* 6=1.5* 2077* T_(1)`

⇒
`T_(1)=(1200000)/(3115.5)`

⇒
`=385.1^(\circ)K` (Initial temperature of helium)

and,

⇒
`pv_(2)=mRT_(2)`

On putting the values, we get

⇒
`200000* 2=1.5* 2077* T_(2)`

⇒
`T_(2)=(400000)/(3115.5)`

⇒
`=128.3^(\circ)K` (Final temperature of helium)