Question

2 mol of an ideal gas expands from 1 L to 10 L reversibly and
isothermally at 250 K. The work done in the process is (R = 2 cal K-1
mol-1)
a) -2.303 kcal
b) 2.303 kcal
c) -5.757 kcal
d) 5.757 kcal

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Give answer with explanation!!
Class 11 thermodynamics.
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Answer 1

Given:

R= 2cal k-1 mol-1

n=2 mol

T= 250K

V1=1L   V2= 10L

Explanation:

W.D.= -nRTln(v2/v1)

so put all values given above,

w.d. = -2*2*250*ln $\frac{10 }{1}$

      = -2303 Cal = -2.303Kcal

PLEASE MARK MY ANSWER AS BRAINLIEST.

Answer 2

OPTION 1)IS CORRECT ANSWER ✅✌