Question

Calculate q, ∆H, and ∆ E when 16 gm of O2 is compressed isothermal and reversibly from 10 litre to 5 litre at 27°C

Answer 1

Answer:

Mass of argon = 10 g

Temperature = 27^{0}C = 27 + 273 = 300 K

Initial volume = 10 L

Final volume = 5 L

Molar mass of argon = 40 g/mole

Value of R (in Cal/mole K) = 2 Cal/mole K

Formula used for Reversible isothermal system is,

For 'q' : q=2.303\times n\times R\times T\times log[\frac{V_{final} }{V_{initial}}] ...........(1)

For ΔH : ΔH = nC_{P}dT ...........(2)

First, we have to calculate the 'moles of argon'

Moles of argon = \frac{\text{ Mass of argon}}{\text{ Molar mass of argon}} = \frac{10g}{40g/mole} = 0.25 moles

Now put all the given values in above formula (1), we get

q=2.303\times 0.25moles\times 2Cal/mol K\times 300K\times log[\frac{5L}{10L}] = - 103.98045 Cal

And the value of ΔH is equal to zero. Because our system is in isothermal, this means system at constant temperature.

The formula of ΔH = nC_{P}dT , and dT =0

Therefore, the value of ΔH become zero.