Question

How much energy is absorbed by 10 mol of
an ideal gas if it expands reversibly from an
initial pressure of 8 atm. to 4 atm at a constant
temperature of 27°C ?
(1) 1.728 x 10^4 J (2) 2 x 10^5 J
(3) 8 x 10^6 J
(4) 1.728 x 10^5 J
plzz provide full solution ​

Answer 1

Answer:

17232.4J

Explanation:

w=-2.303 nrt log(8/4)

w= - 2.303× 10×8.3×300 ×0.3

= -17232.4

work done is absorbed as energy

energy absorbed =17232.4J

Answer 2

$1.728\times 10^4J$ is absorbed by 10 mol of  an ideal gas if it expands reversibly from an  initial pressure of 8 atm. to 4 atm at a constant  temperature of 27°C

Explanation:

The expression used for work done in reversible isothermal expansion will be,

$w=nRT\ln (\frac{P_1}{P_2})$

where,

w = work done = ?

n = number of moles of gas  = 10 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas  = $27^0C=(27+273)K=300K$

$P_1$ = initial pressure of gas  = 8 atm

$P_2$ = final pressure of gas  = 4 atm

Now put all the given values in the above formula, we get:

$w=10mole\times 8.314J/moleK\times 300K\times \ln (\frac{8}{4})$

$w=1.728\times 10^4J$

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