Question

0.5 mole of gas at temp 300 K expands

isothermally from an initial volume of 2.0L

to final volume of 6.0L. (a) What is the work

done by the gas ? (R = 8.31 J mol-1 K-1),

(b) How much heat is supplied to the gas?​

Answer 1

$\huge\underline\frak{\fbox{AnSwEr :-}}$

Step-by-step explanation:

Solution: (a) The work done in isothermal

expansion is W = nRTln

V/Vfi

Where n = 0.5, Vf

= 6L, Vi

= 2L

W = 0.5 mol

8 319 300 62K

.ln mol

K L/L

= 1.369 kJ.

(b) From the first law of thermodynamics,

the heat supplied in an isothermal process

is spent to do work an a system. Therefore,

Q = W = 1.369 kJ.

Answer 2

Given: 0.5 moles of gas at temp 300 K expands

isothermally from an initial volume of 2.0L to a final volume of 6.0L.

To find: We have to find the work done by the gas and heat supplied to the gas.

Solution:

Work done by the gas is given as-

W=nRT ln(V2/V1)

Where n is the number of moles=0.5 moles.

R is the Rydberg constant=8.314

T is the temperature=300K.

V1 is the initial volume= 2L.

V2 is the final volume=6L.

Putting the values in the formula we get-

$w = 0.5 \times 8.314 \times 300 ln( \frac{6}{2} ) \\ w = 1370$

The work done by the gas is 1370J.

For the isothermal process heat supplied is equal to work done.

So, 1370J heat is supplied to the gas.