JEE MAIN - Chemistry (2019 - 12th January Evening Slot - No. 19)
The combination of plots which does not represent isothermal expansion of an ideal gas is –
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A and D
B and D
B and C
A and C
Explanation
For isothermal process of ideal gas,
PV = constants = K
$$ \therefore $$ P = $${k \over v}$$
So, the graph between P and $${1 \over v}$$ is straight line passing through the origin. So, graph (A) is correct.
As PV = K so the P and V curve is hyperbola. So, graph (B) is wrong.
In isothermal PV = constant.
So, graph (C) is right.
We know internal energy (u) = $${f \over 2}$$ nRT. Internal energy is function of temperature (T) only. So, U does not change when volume(V) changes. So, graph (D) is wrong.
PV = constants = K
$$ \therefore $$ P = $${k \over v}$$
So, the graph between P and $${1 \over v}$$ is straight line passing through the origin. So, graph (A) is correct.
As PV = K so the P and V curve is hyperbola. So, graph (B) is wrong.
In isothermal PV = constant.
So, graph (C) is right.
We know internal energy (u) = $${f \over 2}$$ nRT. Internal energy is function of temperature (T) only. So, U does not change when volume(V) changes. So, graph (D) is wrong.
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