Question 1: Thermodynamics & KTG

Question

5.6 liter of helium gas at STP is adiabatically compressed to 0.7 liter. Taking the initial temperature to be
T
1
, the work done in the process is

Accepted Answer

Initially
V
1
= 5.6 l, T
1
= 273 K, P
1
= 1 atm,
$\gamma  = {5 \over 3}$ (For monoatomic gas)
The number of moles of gas is $n = {{5.6l} \over {22.4l}} = {1 \over 4}$
Finally (after adiabatic compression)
V
2
= 0.7 l
For adiabatic compression ${T_1}V_1^{\gamma  - 1} = {T_2}V_2^{\gamma  - 1}$
$	herefore$ ${T_2} = {T_1}{\left( {{{{V_1}} \over {{V_2}}}} ight)^{\gamma  - 1}} = {T_1}{\left( {{{5.6} \over {0.7}}} ight)^{{5 \over 3} - 1}} = {T_1}{(8)^{2/3}} = 4{T_1}$
Work done during an adiabatic process is
$W = {{nR[{T_1} - {T_2}]} \over {(\gamma  - 1)}} = {{{1 \over 4}R[{T_1} - 4{T_1}]} \over {\left[ {{5 \over 3} - 1} ight]}} =  - {9 \over 8}R{T_1}$
Negative sign shows that work is done on the gas.

Answer

${9 \over 8}R{T_1}$

Answer

${3 \over 2}R{T_1}$

Answer

${15 \over 8}R{T_1}$

Answer

${9 \over 2}R{T_1}$

Thermodynamics & KTG

5.6 liter of helium gas at STP is adiabatically compressed to 0.7 liter. Taking the initial temperature to be T 1 , the work done in the process is

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