Question

27°C one mole of an ideal gas is compressed isothermally and reversible from a pressure of 2 atm to 10 atm . the value of delta E and q are (R=2 cal)

Answer 1

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Dear student,
                  given: temperature (T) = 27 0 C =300 K
                                             Δ T=0
                                           P1= 2atm , P2=10atm , n =1 mole
                                           ΔU=? , q=?
For an isothermal and reversible contraction process, we have the following equation for work done 
                           w= 2.303 RT log10V2V1         ------eqn 1
since is done on the system it is positive

By ideal gas equation we know 
     PV=nRT
P1P2=V2V1
Replacing volume in eqn 1 by pressure , we get
w= 2.303 nRTlog​10P1P2              ----eqn 2
substituting all the given values in eqn 2 , we get
w=2.303×1×8.314×300 ×log10(210)
=5744.14×log10(0.2)
= −4014.98 joules or 4.014×103J
= −4.1 KJ
 from the law of thermodynamics for any reversible process
                                 q= −w
so,                           q=−(−4.1) KJ
                              ​ q= 4.1 KJ
 
 ∆U= q+w
       = 4.1-4.1
   ∆U  = 0

Answer 2

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