The pressure P, temperature T, and volume V of one mole of

Question

Problem 8E

The pressure $P$, temperature $T$, and volume $V$ of one mole of an ideal gas are related by the equation PV = 8.31T , when $P$ is measured in kilopascals, $T$ is measured in kelvins, and $V$ is measured in liters.

a. Assume that $P=242.52\pm0.03\mathrm{\ kPa}$ and $V=10.103\pm0.002\mathrm{\ L}$. Estimate $T$, and find the uncertainty in the estimate.

b. Assume that $P=242.52\pm0.03\mathrm{\ kPa}$ and $T=290.11\pm0.02\mathrm{\ K}$. Estimate $V$, and find the uncertainty in the estimate.

c. Assume that $V=10.103\pm0.002\mathrm{\ L}$ and $T=290.11\pm0.02\mathrm{\ K}$. Estimate $P$, and find the uncertainty in the estimate.

Equation Transcription:

$P$

$T$

$V$

$P=242.52\pm{}0.03\ kPa$

$V=10.103\pm{}0.002\ L$

$T=290.11\pm{}0.02\ K$

Text Transcription:

P

T

V

P = 242.52 pm 0.03 kPa

V = 10.103 pm 0.002 L

T = 290.11 pm 0.02 K

Answer 1

Step-by-Step Solution:

Step 1 of 3

Solution 8E

Step1 of 4:

We have Pressure = P, Temperature = T, Volume = V.

One mole of an ideal gas are related by the equation PV = 8.31T.

Here pressure p is measured in kilopascals, T is measured in Kelvins, and V is measured in liters.

Here is our goal is:

a).We need to estimate T, and find the uncertainty in the estimate by assuming that P = 242.52 ± 0.03 kPa and V = 10.103 d = 0.002 L.

b).We need to estimate V, and find the uncertainty in the estimate by assuming that

P = 242.52 ± 0.03 kPa and T = 290.11 ± 0.02 K.

c).We need to Estimate P, and find the uncertainty in the estimate by assuming that V = 10.103 ± 0.002 L and T = 290.11 ± 0.02 K.

Step2 of 4:

a).

From the given information we have P = 242.52, ${\sigma{}}_{P}=0.03$ , V = 10.103 and d = 0.002L.

We have PV = 8.31T

Then,

T = $\frac{PV}{8.31}$

= $\frac{(242.52)(10.103)}{8.31}$

= $\frac{2450.1795}{8.31}$

= 294.8471.

Hence, T = 294.8471.

Consider,

T = $\frac{PV}{8.31}$

Differentiate above equation with respect to “P” then

$\frac{\partial{}T}{\partial{}P}=\frac{\partial{}}{\partial{}P}(\frac{PV}{8.31})$

= $\frac{v}{8.31}\frac{\partial{}}{\partial{}P}(P)$

= $\frac{v}{8.31}$ (1)

Substitute V value in above equation we get

= $\frac{10.103}{8.31}$

= 1.2157

Hence, $\frac{\partial{}T}{\partial{}P}$ = 1.2157.

Again consider,

T = $\frac{PV}{8.31}$

Differentiate above equation with respect to “V” then

$\frac{\partial{}T}{\partial{}V}=\frac{\partial{}}{\partial{}V}(\frac{PV}{8.31})$

= $\frac{P}{8.31}\frac{\partial{}}{\partial{}V}(V)$

= $\frac{P}{8.31}$ (1)

Substitute P value in above equation we get

= $\frac{242.52}{8.31}$

= 29.1841

Hence, $\frac{\partial{}T}{\partial{}V}$ = 29.1841.

The estimate of uncertainty is given by

${\sigma{}}_{T}=\sqrt{{(\frac{\partial{}T}{\partial{}P})}^{2}{\sigma{}}_{P}^{2}+{(\frac{\partial{}T}{\partial{}V})}^{2}{\sigma{}}_{V}^{2}}$

${\sigma{}}_{T}=\sqrt{{(1.2157)}^{2}{(0.03)}_{\ }^{2}+{(29.1841)}^{2}{(0.002)}_{\ }^{2}}$

$=\sqrt{(1.4779)(0.0009)+(851.7116)(0.000004)}$

= $\sqrt{0.001330+0.003406}$

= $\sqrt{0.004736}$

= 0.0688.

Hence, ${\sigma{}}_{T}$ = 0.0688.

Therefore, The estimate of uncertainty is T = 294.8471 $\pm{}$ 0.0688 kelvin.

Step3 of 4

b).

We have P = 242.52, ${\sigma{}}_{P}=0.03,$ T = 290.11, ${\sigma{}}_{T}=0.02$

Consider,

T = $\frac{PV}{8.31}$

V = $\frac{T\times{}8.31}{P}$

= $\frac{290.11(8.31)}{242.52}$

= $\frac{2410.8141}{242.52}$

= 9.9406

Hence, V = 9.9406.

Now,

V = $\frac{T\times{}8.31}{P}$

Differentiate above equation with respect to “P” then

$\frac{\partial{}V}{\partial{}P}$ = $\frac{\partial{}}{\partial{}P}$ ( $\frac{T\times{}8.31}{P})$

= (8.31)T $\frac{\partial{}V}{\partial{}P}(\frac{1}{P})$

= (8.31)T $\frac{\partial{}V}{\partial{}P}$ ( ${P}^{-1}$ )

= (8.31)T(- ${P}^{-1-1}$ )

= -(8.31)T( ${P}^{-2}$ )

= -(8.31)(290.11)( ${242.52)}^{-2}$

= -0.0409

Hence, $\frac{\partial{}V}{\partial{}P}$ = -0.0409.

Consider,

V = $\frac{T\times{}8.31}{P}$

Differentiate above equation with respect to “T” then

$\frac{\partial{}V}{\partial{}T}$ = $\frac{\partial{}}{\partial{}T}$ ( $\frac{T\times{}8.31}{P})$

= ( $\frac{8.31}{P}$ ) $\frac{\partial{}}{\partial{}P}(T)$

= ( $\frac{8.31}{P}$ )(1)

= ( $\frac{8.31}{242.52}$ )

= 0.03426

Hence, $\frac{\partial{}V}{\partial{}T}$ = 0.0324.

The estimate of uncertainty is given by

${\sigma{}}_{T}=\sqrt{{(\frac{\partial{}V}{\partial{}P})}^{2}{\sigma{}}_{P}^{2}+{(\frac{\partial{}V}{\partial{}T})}^{2}{\sigma{}}_{V}^{2}}$

${\sigma{}}_{T}=\sqrt{{(-0.0409)}^{2}{(0.03)}_{\ }^{2}+{(0.0324)}^{2}{(0.02)}_{\ }^{2}}$

$=\sqrt{(0.001628)(0.0009)+(0.00104)(0.0004)}$

= $\sqrt{0.00000146+0.00000056}$

= $\sqrt{0.00000202}$

= 0.001421.

Hence, ${\sigma{}}_{V}$ = 0.001421.

Therefore, The estimate of uncertainty is V = 9.9406 $\pm{}$ 0.001421 liters.

Step4 of 4:

c).

We have V = 10.103, ${\sigma{}}_{V}=0.002,$ T = 290.11, ${\sigma{}}_{T}=0.02$ .

Consider,

T = $\frac{PV}{8.31}$

(8.31)T = PV

P = $\frac{(8.31)T}{V}$

= $\frac{(8.31)(290.11)}{10.103}$

= $\frac{2410.8141}{10.103}$

= 238.6235

Step 2 of 3

Chapter 3.4, Problem 8E is Solved

Step 3 of 3

The pressure P, temperature T, and volume V of one mole of

Solution for problem 8E Chapter 3.4

Chapter 3.4, Problem 8E is Solved

Textbook: Statistics for Engineers and Scientists

Edition: 4

Author: William Navidi

ISBN: 9780073401331

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