5.6 liter of Helium gas at STP is adiabatically compressed to 0.7 liter. Taking the initial temperature to be T1, then work done on the system in the process is (no. of moles =1 )

Question

5.6 liter of Helium gas at STP is adiabatically compressed to 0.7 liter. Taking the initial temperature to be T1, then work done on the system in the process is (no. of moles =1 ) (A) (9/8)RT1 (B) (3/2)RT1 (C) (15/8)RT1 (D) (9/2)RT1

Tardigrade Admin · Accepted Answer

T1V1&gamma; - 1=T2V2&gamma; - 1 Ti⋅ V7 / 5 - 1=T2⋅ ((1/32) V)7 / 5 - 1 T2=4Ti=Î±Ti α =4

Q. $5.6$ liter of Helium gas at $STP$ is adiabatically compressed to $0.7$ liter. Taking the initial temperature to be $T_{1},$ then work done on the system in the process is (no. of moles $= 1$ )

$\frac{9}{8} R T_{1}$

$\frac{3}{2} R T_{1}$

$\frac{15}{8} R T_{1}$

$\frac{9}{2} R T_{1}$

$T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1}$
$T_{i} \cdot V^{7/5 - 1} = T_{2} \cdot (\frac{1}{32} V)^{7/5 - 1}$
$T_{2} = 4 T_{i} = α T_{i}$
$α = 4$

Q. 5.6 liter of Helium gas at STP is adiabatically compressed to 0.7 liter. Taking the initial temperature to be T1​, then work done on the system in the process is (no. of moles =1 )

89​RT1​

23​RT1​

815​RT1​

29​RT1​