Question

20 × 10-3 kg of argon is allowed to expand isothermally and reversibly from 11dm3 to certain volume and heat absorbed is 2.78 kj at 300 k. Calculate the final volume.

Answer 1

Answer : The final volume of argon gas will be, 102 liters

Explanation :

This is the case of isothermal reversible expansion of gas.

Formula used :

$q=2.303nRT\log (\frac{V_2}{V_1})\\\\q=2.303\times \frac{m}{M}\times RT\log (\frac{V_2}{V_1})$

where,

q = heat absorbed by the gas  = 2.78 KJ = 2780 J

n = number of moles of gas

R = gas constant = 8.314 J/mole K

T = temperature of gas = 300 K

m = mass of argon gas = $20\times 10^{-3}Kg=20g$

M = molar mass of argon gas = 40 g/mole

$V_1$ = initial volume of gas = $11dm^3=11L$

$V_2$ = final volume of gas = ?

Now put all the given values in the above formula, we get the final volume of the gas.

$2780J=2.303\times \frac{20g}{40g/mole}\times 8.314J/moleK\times 300K\log (\frac{V_2}{11L})$

$V_2=102L$

Therefore, the final volume of argon gas will be, 102 liters