Which of the following graphs correctly represents the variation of with for an ideal gas at constant temperature? [2002]

Question

Which of the following graphs correctly represents the variation of $β = - \frac{d V / d P}{V}$ with $P$ for an ideal gas at constant temperature? [2002]

Smart Achievers · Accepted Answer

(1)

$According to Boyle's law, P V = constant.$

$∴$ $P d V + V d P = 0$

$\Rightarrow \frac{P d V}{d P} = - V$ $; β = - (\frac{1}{V}) (\frac{d V}{d P}) = (\frac{1}{P})$

$\Rightarrow β \times P = 1$

$Hence graph between β and P will be a rectangular hyperbola.$

Solution

Q.
Which of the following graphs correctly represents the variation of $β = - \frac{d V / d P}{V}$ with $P$ for an ideal gas at constant temperature [2002]

1

2

3

4

Ans.
(1)

$According to Boyle's law, P V = constant.$

$∴$ $P d V + V d P = 0$

$\Rightarrow \frac{P d V}{d P} = - V$ $; β = - (\frac{1}{V}) (\frac{d V}{d P}) = (\frac{1}{P})$

$\Rightarrow β \times P = 1$

$Hence graph between β and P will be a rectangular hyperbola.$

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Solution

Q. Which of the following graphs correctly represents the variation of β=-dV/dPV with P for an ideal gas at constant temperature [2002]

1

2

3

4

Ans. (1) According to Boyle's law, PV=constant. ∴ PdV+VdP=0 ⇒PdVdP=-V; β=-(1V)(dVdP)=(1P) ⇒β×P=1 Hence graph between β and P will be a rectangular hyperbola.

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Q.
Which of the following graphs correctly represents the variation of $β = - \frac{d V / d P}{V}$ with $P$ for an ideal gas at constant temperature [2002]

Ans.
(1)

$According to Boyle's law, P V = constant.$

$∴$ $P d V + V d P = 0$

$\Rightarrow \frac{P d V}{d P} = - V$ $; β = - (\frac{1}{V}) (\frac{d V}{d P}) = (\frac{1}{P})$

$\Rightarrow β \times P = 1$

$Hence graph between β and P will be a rectangular hyperbola.$