The value of PcVc-RTc is 3-x-value of x is-

V_C=3b #160; #160;P_C=127b^2 #160; #160; and #160; #160; T_C=8a27Rb P_CV_CRT_C=a27b^2× 3bR×8b27Rb = a9b8Ra27Rb=a9b8a27b ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Liquefaction of Gases and Critical Constants

The value of ...

Question

The value of $\frac{P_{c} V_{c}}{R T_{c}}$ is $3 / x$ .
value of $x$ is............

Open in App

Solution

$V_{C} = 3 b P_{C} = \frac{1}{27 b^{2}}$ and $T_{C} = \frac{8 a}{27 R b}$

$\frac{P_{C} V_{C}}{R T_{C}} = \frac{\frac{a}{27 b^{2}} \times 3 b}{R \times \frac{8 b}{27 R b}}$ = $\frac{\frac{a}{9 b}}{\frac{8 R a}{27 R b}} = \frac{\frac{a}{9 b}}{\frac{8 a}{27 b}}$

$\frac{P_{C} V_{C}}{R T_{C}} = \frac{a}{9 b} \times \frac{27 b}{8 a} = \frac{3}{8} = 0.375$

Suggest Corrections

Similar questions

Q. The value of

\frac{P_{c} V_{c}}{R T_{c}}

Q. Calculate the volume occupied by 14.0 g

N_{2}

at 200 K and 8.21 atm pressure if

\frac{P_{c} V_{c}}{R T_{c}} = \frac{3}{8}

and

\frac{P_{r} V_{r}}{T_{t}} = 2.2

[write the value to the nearest integer]

Q. Calculate the volume occupied by

2.0 m o l e

N_{2}

200 K

and

8.21 a t m

pressure, if

\frac{P_{C} V_{C}}{R T_{C}} = \frac{3}{8}

and

\frac{P_{r} V_{r}}{T_{r}} = 2.4

Q. Calculate the volume occupied by

2.0

mole of

N_{2}

200

K and

8.21

atm pressure, if

\frac{P_{C} V_{C}}{R T_{C}} = \frac{3}{8}

and

\frac{P_{r} V_{r}}{T_{r}} = 2.4

The volume occupied by 2 mol of $N_{2}$ at 200 K and 10.1325 MPa pressure is

(Given that $\frac{P_{c} V_{C}}{R T_{C}} = \frac{3}{8}$ and $\frac{P_{r} V_{r}}{T_{r}} = 2.21$ ), where $P_{r}, V_{r}$ and $T_{r}$ are reduced pressure, reduced volume and reduced temperature.

Join BYJU'S Learning Program

The value of PcVcRTc is 3/x.value of x is............

VC=3bPC=127b2 and TC=8a27RbPCVCRTC=a27b2×3bR×8b27Rb=a9b8Ra27Rb=a9b8a27bPCVCRTC=a9b×27b8a=38=0.375

The value of $\frac{P_{c} V_{c}}{R T_{c}}$ is $3 / x$ .
value of $x$ is............