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Physics Question on Thermodynamics

5.6 L of helium gas at STP is adiabatically compressed to 0.7 L. Taking the initial temperature to be T1T_1, the work done in the process is

A

98RT1\frac{9}{8}RT_1

B

32RT1\frac{3}{2}RT_1

C

158RT1\frac{15}{8}RT_1

D

92RT1\frac{9}{2}RT_1

Answer

98RT1\frac{9}{8}RT_1

Explanation

Solution

At STP, 22.4 L of any gas is 1 mole.
∴    5.6L=5.622.4=14moles=n\therefore \, \, \, \, 5.6L =\frac{5.6}{22.4}=\frac{1}{4} moles = n
In adiabatic process
           TVγ−1=constant\, \, \, \, \, \, \, \, \, \, \, TV^{\gamma-1}=constant
∴     T2V2γ−1=T1V1γ−1  or  T2=T−1(V1V2)γ−1\therefore \, \, \, \, \, T_2V_2^{\gamma -1}=T_1V_1^{\gamma-1} \, \, or \, \, T_2=T-1\bigg(\frac{V_1}{V_2}\bigg)^{\gamma-1}
       γ=CpCv=53\, \, \, \, \, \, \, \gamma=\frac{C_p}{C_v}=\frac{5}{3} for monoatomic He gas
∴   T2=T1(5.60.7)53−1 =4T1\therefore \, \, \, T_2=T_1 \bigg(\frac{5.6}{0.7}\bigg)^{\frac{5}{3}-1} \, =4 T_1
Further in adiabatic process, Q = 0
∴         W+ΔU=0\therefore \, \, \, \, \, \, \, \, \, W+\Delta U=0
or     w=−ΔU=−nCvΔT=−n(Rγ−1)(T2−T1) \, \, \, \, w=-\Delta U=-nC_v \Delta T =-n\bigg(\frac{R}{\gamma-1}\bigg)(T_2-T_1)
=-14(R53−1)(4T1−T1)=−98RT1\frac{1}{4}\bigg(\frac{R}{\frac{5}{3}-1}\bigg)(4T_1-T_1)=-\frac{9}{8}RT_1
∴ \therefore \, Correct option is (a)