Question

0.5kg of an ideal gas expands adiabatically until its pressure is halfed. During expansion the gas *does* 30kj of external work and its temprature falls from 500 k to 410 k. Make calculation for the adiabatic exponent and the characteristics gas constant(R). Also find the change in internal energy during adiabatic expansion .​

Answer 1

Answer:

During expansion the gas does 30Kj of external work and its temperature falls from 500K to 410K. Make calculation for the adiabatic

Explanation:

During expansion the gas does 30Kj of external work and its temperature falls from 500K to 410K. Make calculation for the adiabatic.

Answer 2

Answer: Adiabatic exponent of gas is 1.4 and change in internal energy is -30 kJ.

Explanation:

Let initial pressure be $P_o$ and initial volume be $V_o$  and the adiabatic exponent be γ

Assume number of moles to be n

Final pressure =  $\frac{P_o}{2}$

Final volume = V

$P_o$ $V_o$  = nR(500)   eq (i)

$\frac{P_o}{2}$ V = nR(410)     eq (ii)

Divide  eq (ii) by eq (i)

$\frac{V}{2V_o} = \frac{41}{50}$

V  =  $\frac{41}{25}$ $V_o$

In adiabatic process, P(V^γ) = constant

Therefore,

⇒ $P_o$ ($V_o$)^γ  =   ( $\frac{P_o}{2}$ ) ( $\frac{41}{25}$ $V_o$)^γ

⇒  2 = (41/25)^γ

⇒  ㏑(2) = γ㏑($\frac{41}{25}$)

⇒   γ  = 1.4

First law of Thermodynamics :

dq = dU - W

dq ⇒ Heat supplied

dU ⇒ Change in internal energy

W ⇒ Work done on the system

In adiabatic process dq = 0

dU = W

W = - 3O kJ (negative sign because gas is doing the work)

dU =  - 30 kJ

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