JEE MAIN - Physics (2021 - 25th February Evening Shift - No. 10)
Thermodynamic process is shown below on a P-V diagram for one mole of an ideal gas. If V2 = 2V1 then the ratio of temperature T2/T1 is :
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$$\sqrt 2 $$
$${1 \over {\sqrt 2 }}$$
$${1 \over 2}$$
2
Explanation
From P-V diagram,
Given PV1/2 = constant ...... (1)
We know that
PV = nRT
$$P \propto \left( {{T \over V}} \right)$$
Put in equation (1)
$$\left( {{T \over V}} \right){(V)^{1/2}}$$ = constant
$$T \propto {V^{1/2}}$$
$$ \Rightarrow $$ $${{{T_2}} \over {{T_1}}} = \sqrt {{{{V_2}} \over {{V_1}}}} $$
$$ \Rightarrow $$ $${{{T_2}} \over {{T_1}}} = \sqrt {{{2{V_1}} \over {{V_1}}}} $$ [As V2 = 2V1 ]
$$ \Rightarrow $$ $${{{T_2}} \over {{T_1}}} = \sqrt 2 $$
Given PV1/2 = constant ...... (1)
We know that
PV = nRT
$$P \propto \left( {{T \over V}} \right)$$
Put in equation (1)
$$\left( {{T \over V}} \right){(V)^{1/2}}$$ = constant
$$T \propto {V^{1/2}}$$
$$ \Rightarrow $$ $${{{T_2}} \over {{T_1}}} = \sqrt {{{{V_2}} \over {{V_1}}}} $$
$$ \Rightarrow $$ $${{{T_2}} \over {{T_1}}} = \sqrt {{{2{V_1}} \over {{V_1}}}} $$ [As V2 = 2V1 ]
$$ \Rightarrow $$ $${{{T_2}} \over {{T_1}}} = \sqrt 2 $$
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