Question 152: Thermodynamics & KTG

Question

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

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