Question

5 moles of helium expand isothermally and reversibly from a pressure 40 x 10
sibly from a pressure 40 x 10 Nm to 4x10 Noi at 300 K (3)
10 4x
Calculate the work done. Change in internal energy and heat obsorbed during the expansion
(RE8.314 JK 'mo!')​

Answer 1

Answer: The value of W, Q and dU for the given process are -28720.7 J, -28720.7 J and 0 J respectively.

Explanation:

To calculate the work done for isothermal, reversible expansion process, we use the equation:

$W=-2.303nRT\log(\frac{P_1}{P_2})$

where,

W = work done

n = number of moles = 5 mole

R = Gas constant = 8.314 J/mol.K

T = Temperature of the gas = 300 K

$P_1$ = initial pressure = $(40\times 10)Nm=400Nm$

$P_2$ = final pressure =  $(4\times 10)Nm=40Nm$

Putting values in above equation, we get:

$W=-2.303\times 5mol\times 8.314J/mol.K\times 300K\times \log(\frac{400}{40})\\\\W=-28720.7J$

The equation for first law of thermodynamics follows:

$Q=dU+W$

where,

Q = total amount of heat released = ?

dU = Change in internal energy = 0 (for isothermal process)

W = work done

So, Q = W = -28720.7 J

Hence, the value of W, Q and dU for the given process are -28720.7 J, -28720.7 J and 0 J respectively.