At 405 K- 70 ml of one mol of ammonia has dP-dV- 0- then which of the following options is not correct-

DP/dV=0 represents the critical point of ammonia. So, T = T_c= 132^∘ C, V = V_c= 70 ml/mol= 0.07 L/mol T_b = a/Rb, b = V_c/3, a=3P_cV_c^2, From the above, op ...

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Standard XII

Chemistry

Liquefaction of Gases and Critical Constants

At 405 K, 70 ...

Question

At 405 K, 70 ml of one mol of ammonia has $\frac{d P}{d V} = 0$ , then which of the following options is not correct?

T_{c} = 132^{\circ} C

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T_{b} = 1094^{\circ} C

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V_{c} = 0.07 L / m o l

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b = 70 m l / m o l

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Solution

The correct option is D $b = 70 m l / m o l$
$\frac{d P}{d V} = 0$ represents the critical point of ammonia.
So, $T = T_{c} = 132^{\circ} C$ , $V = V_{c} = 70 m l / m o l = 0.07 L / m o l$
$T_{b} = \frac{a}{R b}, b = \frac{V_{c}}{3}, a = 3 P_{c} V_{c}^{2},$
From the above, option d is clearly incorrect.
Also, $T_{b} = \frac{9 P_{c} V_{c}}{R}$
$Z = \frac{P_{c} V_{c}}{R T_{c}} = \frac{3}{8}$
So, $T_{b} = \frac{27}{8} T_{c} \approx 1377 K = 1094^{\circ} C$

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At 405 K, 70 ml of one mol of ammonia has dPdV=0, then which of the following options is not correct?

The correct option is D b=70 ml/moldPdV=0 represents the critical point of ammonia. So, T=Tc=132∘ C, V=Vc=70 ml/mol=0.07 L/mol Tb=aRb,b=Vc3,a=3PcV2c, From the above, option d is clearly incorrect. Also, Tb=9PcVcR Z=PcVcRTc=38 So, Tb=278Tc≈1377 K=1094∘ C

At 405 K, 70 ml of one mol of ammonia has $\frac{d P}{d V} = 0$ , then which of the following options is not correct?