For a reversible process at T=300k, the volume of 2 mole of ideal gas is increased from 1 litre to 10 litres, the $\delta H$ for isothermal change is: Option 1) 11.47 kJ Option 2) 4.98 kJ Option 3) 0 Option 4) 2.49 kJ

As learnt in

First law of Thermodynamics -

Energy of universe is always conserved or total energy of an isolated system is always conserved

$\Delta E= q + W$

- wherein

$\Delta E=$ Internal Energy

$q=$ Heat

$W=$ work

and

Reversible Isobaric Process -

$W=P\left ( V_{f}-V_{i} \right )$

- wherein

$P_{gas}= constant$

Since pressure of gas is defined during whole of the process and hence the process must be reversible

For an isothermal process, $\Delta E =0$

$\Delta H =\Delta E+P\Delta V$

$\Delta H =P\Delta V = nR\Delta T\ \ \ \ \ \left ( \Delta T=0 \right )$

$\therefore \Delta H = 0$

Option 1)

11.47 kJ

Option 2)

4.98 kJ

Option 3)

0

Option 4)

2.49 kJ

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