5.6 ~L of helium gas at STP is adiabatically compressed to 0.7 ~L . Taking the initial temperature to be T₁ , the work done in the process is

Step-by-step solution

At STP, 22.4 L of any gas is 1 mole. 5.6 ~L = 5.6 22.4 = 1 4 moles =n In adiabatic process, array rlrl & T V^ -1 & = constant & & T₂ V₂^ -1 & =T₁ V₁^ -1 & or & T₂ & =T₁ ( V₁ V₂ )^ -1 array aligned & = C_p C_V = 5 3 for monoatomic He gas. & T₂=T₁ ( 5.6 0.7 )^ 5 3 -1 =4 T₁ & Further in adiabatic process, & Q=0 & W+ U=0 & or W=- U & =-n C_V T & =-n ( R -1 ) (T₂-T₁ ) & =- 1 4 ( R 5 3 -1 ) (4 T₁-T₁ ) & =- 9 8 R T₁ & aligned Correct option is (a). Analysis of Question (i) From calculation point of view question is modera