Question 483: Thermodynamics & KTG

Question

One mole of an ideal monoatomic gas is taken along the path ABCA as show in the PV diagram. The maximum temperature attained by the gas along the path BC is  given by :

Accepted Answer

Equation of line BC,
P = P
0
$$-$$ $${{2{P_0}} \over {{V_0}}}\left( {V - 2{V_0}} ight)$$
As $$\,\,\,\,\,$$ PV = nRT
$$	herefore\,\,\,\,$$ T = $${{PV} \over {nR}}$$
= $${{{P_0}V - {{2{P_0}{V_2}} \over {{V_0}}} + 4{P_0}V} \over {1 	imes R}}$$
(As  given n = 1 mole gas)
$$ \Rightarrow $$$$\,\,\,$$ T = $${{{P_0}} \over R}\left[ {5V - {{2{V^2}} \over {{V_0}}}} ight]$$
For maximum value of T
$${{dT} \over {dV}} = 0$$
$$	herefore\,\,\,$$ 5 $$-$$ $${{4V} \over {{V_0}}} = 0$$
$$ \Rightarrow $$$$\,\,\,$$ V = $${5 \over 4}{V_0}$$
$$	herefore\,\,\,$$ T
max
= $${{{P_0}} \over R}\left[ {5 	imes {{5{V_0}} \over 4} - {2 \over {{V_0}}} 	imes {{25} \over {16}}{V_0}^2} ight]$$
= $${{25} \over 8} 	imes {{{P_0}\,{V_0}} \over R}$$

Answer

$${{25} \over {16}}\,{{{P_o}{V_o}} \over R}$$

Answer

$${{25} \over {8}}\,{{{P_o}{V_o}} \over R}$$

Answer

$${{25} \over {4}}\,{{{P_o}{V_o}} \over R}$$

Answer

$${{5} \over {8}}\,{{{P_o}{V_o}} \over R}$$

Thermodynamics & KTG

One mole of an ideal monoatomic gas is taken along the path ABCA as show in the PV diagram. The maximum temperature attained by the gas along the path BC is given by :

Choose the correct answer: