Question

find work done when 1 mole of hydrogen expands isothermally from 15 to 50 litres agianst a constant pressure of 1atm at 25c

Answer 1

P1=4atm = 4*10^5 Pa

V1= 1L = 10^-3 m^3

P2= 1atm

We could use

PV=nRT(ideal gas equation)

P1V1 = nRT1 (T1=T2 given..isothermal) (n=1mole)

4*10^5* 1 = 1* .0821 * T

T= 4*10^5/0.0821 K

W=2.303 nRT * log P1/P2

= 2.303 *1 *.0821 * 4* 10^5/0.0821 * log 4/1

on solving u will get a value of 552 J

Answer 2

P1=4atm = 4*10^5 Pa

V1= 1L = 10^-3 m^3

P2= 1atm

We could use

PV=nRT(ideal gas equation)

P1V1 = nRT1 (T1=T2 given..isothermal) (n=1mole)

4*10^5* 1 = 1* .0821 * T

T= 4*10^5/0.0821 K

W=2.303 nRT * log P1/P2

= 2.303 *1 *.0821 * 4* 10^5/0.0821 * log 4/1

on solving u will get a value of 552 J