Question

As a result of the isobaric heating by DeltaT=72K, one mole of a certain ideal gas obtain an amount of heat Q=1.6kJ. Find the work performed by the gas, the increment of its internal energy and gamma.

Answer 1

Given :

For isobaric heating , ΔT = 72 kelvin

Number of mole = n = 1

Amount of heat obtain = Q = 1.6 k Joule

To Find :

The work performed by the gas

The value of gamma = $\gamma$

Solution :

Let The work done = W joule

∵  work done = $\int P dV$

                     = P $\int dV$

                      = P ΔV

And ∵  P V = n R T              (ideal gas law )

where    n = number of moles

And       R = 8.314

So, work done = W = n R ΔT  

And for n = 1 mole

W = 1 × 8.314 × ΔT

i.e   W = 8.314 × 72

∴   Work done = W = 598.6   joule  = 0.598 k J

Again

From first law of thermodynamics

      ΔQ = ΔU + ΔW

Or,    ΔU = ΔQ - ΔW

Or,    ΔU = 1.6  k J - 0.598 K J

∴      ΔU = 1.002  k J  ≈ 1 k J

So, The change in internal energy =  ΔU  = 1 k J

Now,

∵      ΔU = n $c_v$  ΔT

And   ΔQ = n  $c_p$  ΔT

Since we know that ,   $\gamma$  = $\dfrac{c_p}{c_v}$

So,     $\dfrac{\Delta Q}{\Delta U}$  = $\dfrac{n c_p\Delta T}{n c_v\Delta T}$

Or,    $\dfrac{1.6 KJ}{1K J}$ = $\dfrac{c_p}{c_v}$

I.e     $\gamma$  = $\dfrac{c_p}{c_v}$ = 1.6

So, The value of $\gamma$ = 1.6

Hence, The change in internal energy is 1 k J   and The value of gamma is 1.6  Answer

Answer 2

Answer:

Explanation:

Given :

For isobaric heating , ΔT = 72 kelvin

Number of mole = n = 1

Amount of heat obtain = Q = 1.6 k Joule

To Find :

The work performed by the gas

The value of gamma =  

Solution :

Let The work done = W joule

∵  work done =  

                    = P  

                     = P ΔV

And ∵  P V = n R T              (ideal gas law )

where    n = number of moles

And       R = 8.314

So, work done = W = n R ΔT  

And for n = 1 mole

W = 1 × 8.314 × ΔT

i.e   W = 8.314 × 72

∴   Work done = W = 598.6   joule  = 0.598 k J

Again

From first law of thermodynamics

     ΔQ = ΔU + ΔW

Or,    ΔU = ΔQ - ΔW

Or,    ΔU = 1.6  k J - 0.598 K J

∴      ΔU = 1.002  k J  ≈ 1 k J

So, The change in internal energy =  ΔU  = 1 k J

Now,

∵      ΔU = n   ΔT

And   ΔQ = n    ΔT

Since we know that ,     =  

So,       =  

Or,     =  

I.e       =  = 1.6

So, The value of  = 1.6

Hence, The change in internal energy is 1 k J   and The value of gamma is 1.6  AnswerGiven :

For isobaric heating , ΔT = 72 kelvin

Number of mole = n = 1

Amount of heat obtain = Q = 1.6 k Joule

To Find :

The work performed by the gas

The value of gamma =  

Solution :

Let The work done = W joule

∵  work done =  

                    = P  

                     = P ΔV

And ∵  P V = n R T              (ideal gas law )

where    n = number of moles

And       R = 8.314

So, work done = W = n R ΔT  

And for n = 1 mole

W = 1 × 8.314 × ΔT

i.e   W = 8.314 × 72

∴   Work done = W = 598.6   joule  = 0.598 k J

Again

From first law of thermodynamics

     ΔQ = ΔU + ΔW

Or,    ΔU = ΔQ - ΔW

Or,    ΔU = 1.6  k J - 0.598 K J

∴      ΔU = 1.002  k J  ≈ 1 k J

So, The change in internal energy =  ΔU  = 1 k J

Now,

∵      ΔU = n   ΔT

And   ΔQ = n    ΔT

Since we know that ,     =  

So,       =  

Or,     =  

I.e       =  = 1.6

So, The value of  = 1.6

Hence, The change in internal energy is 1 k J   and The value of gamma is 1.6  Answer