Question

Calculate w,q and ∆V when 0.75 mol of an ideal gas exapands isothermally and reversibly at 27°c from a volume of 15L to 25L

Answer 1

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Answer 2

a) w = -955.6 J

b) q= 955.6 J

c) $\Delta V$ = 10 L

Explanation:

Formula used for isothermal reversible ideal gas is,

$w=-nRT\ln (\frac{V_2}{V_1})$

where,

w = work done

n = number of moles of gas = 0.75 mole

R = gas constant = 8.314 J/moleK

T = temperature of gas = $27^0C=(27+273)K=300 K$

$V_1$ = initial volume of gas = 15 L

$V_2$ = final volume of gas = 25 L

Now put all the given values in the above formula, we get the work done.

$w=-[(0.75mole)\times (8.314J/moleK)\times (300K)]\ln (\frac{25}{15})$

$w=-955.6J$

According to first law of thermodynamics:

$\Delta E=q+w$

$\Delta E$=Change in internal energy  = 0 (for isothermal reaction as internal energy is a function of temperature)

q = heat absorbed or released  = ?

w = work done or by the system  = -955.6J

$0=q+(-955.6)J$

$q=955.6J$

$\Delta V$ =change in volume= $V_2-V_1=(25-15)L=10L$

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