One mole of an ideal monatomic gas initially at temperature 300 K is isothermally expanded to four times its initial volume and then isochorically heated so that the pressure in the final state becomes equal to its initial value. The process is shown in the P-V graph (Take, $R= \frac{25}{3} {J mol}^{-1} k^{-1}$ )

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Slope of the curve $A \to B$ when volume is ${50m}^{3}$ is –5 in SI units

Heat absorbed in the process $A \to B \to C$ is nearly 11.25 KJ

Heat absorbed in the process $A \to B \to C$ is nearly 10.7 KJ

Slope of the curve $A \to B$ when volume is ${50m}^{3}$ is –1 in SI units

answer is B.

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Detailed Solution

Apply first law of thermodynamics
$P V = R T$
$\Rightarrow P = \frac{8.3 \times 300}{25} = 100$
Slope $= \frac{- P}{V} = - \frac{100}{2 \times 50} = - 1 Sl unit$

$Δ Q = \frac{25}{3} \times 300 In 4 + \frac{25}{3} \times \frac{3}{2} \times 900$
$= 5000 \times 0.69 + 11250 = 14700 J$