Question

Calculate w,q,and du when 0.75 mol of an ideal gas expand isothermally and reversible at 27°c from vol. 15L to 25L

Answer 1

For an isothermal reversible expansion of an ideal gas

Putting n = 0.75 mol; V1 = 15 L; V2 = 25 L, T = 27 + 273 = 300 K

R = 8.314 JK–1 mol–1

Answer 2

Answer: The value of W, Q and dU for the given process are -955.75 J, -955.75 J and 0 J respectively.

Explanation:

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

$W=-2.303nRT\log(\frac{V_2}{V_1})$

where,

W = work done

n = number of moles = 0.75 mole

R = Gas constant = 8.314 J/mol.K

T = Temperature of the gas = $27^oC=[27+273]K=300K$

$V_1$ = initial volume = 15 L

$V_2$ = final volume = 25 L

Putting values in above equation, we get:

$W=-2.303\times 0.75mol\times 8.314J/mol.K\times 300K\times \log(\frac{25}{15})\\\\W=-955.75J$

The equation for first law of thermodynamics follows:

$Q=dU+W$

where,

Q = total amount of heat required = ?

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

W = work done

So, Q = W = -955.75 J

Hence, the value of W, Q and dU for the given process are -955.75 J, -955.75 J and 0 J respectively.