A gas at 350 K 15 atm has a molar volume 12 percent smaller than that calculated from the perfect gas law. Compressibility factor under these conditions can be expressed in scientific notation as 88 × 10-x. The value of x is

Question

A gas at 350 K 15 atm has a molar volume 12 percent smaller than that calculated from the perfect gas law. Compressibility factor under these conditions can be expressed in scientific notation as 88 × 10-x. The value of x is  (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Z=(P V m /R T)=(V m /V m text perfect ) Given, V m =V m perfect -0.12 V m perfect =0.88 V m perfect Z =( V m / V m textperfect )=0.88 =88 × 10-2

Q. A gas at $350 K 15 a t m$ has a molar volume $12$ percent smaller than that calculated from the perfect gas law. Compressibility factor under these conditions can be expressed in scientific notation as $88 \times 1 0^{- x}$ . The value of $x$ is ___

$Z = \frac{P V _{m}}{RT} = \frac{V _{m}}{V _{m} perfect}$
Given, $V_{m} = V_{m}$ perfect $- 0.12 V_{m}$ perfect
$= 0.88 V_{m}$ perfect
$Z = \frac{V _{m}}{V _{m} perfect} = 0.88$
$= 88 \times 1 0^{- 2}$

Q. A gas at 350K15atm has a molar volume 12 percent smaller than that calculated from the perfect gas law. Compressibility factor under these conditions can be expressed in scientific notation as 88×10−x. The value of x is ___

Z=RTPVm​​=Vm​ perfect Vm​​ Given, Vm​=Vm​ perfect −0.12Vm​ perfect =0.88Vm​ perfect Z=Vm​perfectVm​​=0.88 =88×10−2

Q. A gas at $350 K 15 a t m$ has a molar volume $12$ percent smaller than that calculated from the perfect gas law. Compressibility factor under these conditions can be expressed in scientific notation as $88 \times 1 0^{- x}$ . The value of $x$ is ___

$Z = \frac{P V _{m}}{RT} = \frac{V _{m}}{V _{m} perfect}$
Given, $V_{m} = V_{m}$ perfect $- 0.12 V_{m}$ perfect
$= 0.88 V_{m}$ perfect
$Z = \frac{V _{m}}{V _{m} perfect} = 0.88$
$= 88 \times 1 0^{- 2}$