find work done when 1 mole of hydrogen expands isothermally from 15 to 50 litres agianst a constant pressure of 1atm at 25c
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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
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
Answered by
0
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
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