Question

Solve
Q.11. 10 g of argon gas is compressed isothermally and reversibly at a temperature of 27°C from 10 L to 5 L. Calculate q and ∆H for this process. Atomic wt. of Ar = 40.
1) + 103.99 cal, 0
3) - 39.99 cal, 3 J
2) + 39.99 cal, 3 J
4) - 103.99 cal, 0

Answer 1

Answer : (4) - 103.99 Cal , 0 J

Solution : Given,

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.

Answer 2

Answer:

-103.99 Cal , 0

Explanation:

mass of argon is 10 grams

temperature=27°C=300K

molar mass of argon is 40 gram/mole

R=2 calorie/mole Kelvin

number of moles of argon = mass of argon/molar mass of argon

number of moles of argon=10/40=0.25 moles

q =-2.303×nRT×logv1/v2

q = -2.303×0.25×2×300×log1/2

q =-103.99

Delta H = 0 because the system is in isothermal