Question

Calculate the maximum work when 24 g of oxygen are expanded isothermally and reversibly from a pressure of 1.6 x 10⁵ Pa to 100 kPa at 298 K. (-873.4 J)

Answer 1

Answer:

W.d = -873.4 J

Explanation:

Since we know that the work done for the isothermal process $w.d = -nRTln\dfrac{P_{1} }{P_{2}}$................(1)

Isothermal process is the process in which temperature remains constant.

Where n is no.of moles

$n = \dfrac{24}{32} = \dfrac{3}{4}$

R = 8.314 ,T = 298 K.

$P_{1} = 1.6\times 10^{5}\ pa= 160\ kpa$

$P_{2} = 100\ kpa$

Using (1) equation we get,

$W.d = -\dfrac{3}{4}\times 8.314\times 298\ ln\dfrac{160}{100} = -873.4\ J$

Answer 2

Explanation:

Solution:-

Given weight =24g

Molecular weight =32g

No. of moles (n)=3224=0.75 mol

As we know that work done in an isothermal process is given as-

W=−2.303nRTlogP2P1

Given:-

T=298K

P1=1.6×105Pa=160kPa

P2=100kPa

Therefore,

W=−2.303×0.75×0.0821×298×log100160

⇒W=−42.26×log1.6=−8.62atm−L