Question

16 kg oxygen gas expands at STP (1 atm)
isobarically to occupy double of its original volume.
The work done during the process is nearly
(1) 260 kcal (2) 180 kcal
(3) 130 kcal
(4) 271 kcal​

Answer 1

Answer:

271

Explanation:

Given 16 kg oxygen gas expands at STP (1 atm)

isobarically to occupy double of its original volume.

The work done during the process is nearly

We know that

     W = - p (Δ v)

         = - 1 atm (v2 – v1)

Now we need to find v1 and v2

So 16 kg = 16,000 gms

So v1 = 22.4 x 16,000

         = 3,58,400 / 32

        = 11,200

Now v2 is double the original volume.

So v2 = 2 x 11,200

         = 22,400

Now w = -1(22,400 – 11,200)

   So w = -11200

           = 11,200 litre atmosphere

We know that 1 L atm = 101.3 J

So 11200 x 101.3 = 11,34,560 J

We know that  

1 J = 0.000239 K cal

So 11,34560 J will be equal to 11,34,560 x 0.000239 = 271 K cal

Answer 2

Answer:

Answer:

271

Explanation:

Given 16 kg oxygen gas expands at STP (1 atm)

isobarically to occupy double of its original volume.

The work done during the process is nearly

We know that

    W = - p (Δ v)

        = - 1 atm (v2 – v1)

Now we need to find v1 and v2

So 16 kg = 16,000 gms

So v1 = 22.4 x 16,000

        = 3,58,400 / 32

       = 11,200

Now v2 is double the original volume.

So v2 = 2 x 11,200

        = 22,400

Now w = -1(22,400 – 11,200)

  So w = -11200

          = 11,200 litre atmosphere

We know that 1 L atm = 101.3 J

So 11200 x 101.3 = 11,34,560 J

We know that  

1 J = 0.000239 K cal

So 11,34560 J will be equal to 11,34,560 x 0.000239 = 271 K cal

Explanation: