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

ISBN: 9780073401331 38

## Solution for problem 8E Chapter 3.4

Statistics for Engineers and Scientists | 4th Edition

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Statistics for Engineers and Scientists | 4th Edition

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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:

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

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, , V = 10.103 and d = 0.002L.

We have PV = 8.31T

Then,

T =

=

=

= 294.8471.

Hence, T = 294.8471.

Consider,

T =

Differentiate above equation with respect to “P” then

=

= (1)

Substitute V value in above equation we get

=

= 1.2157

Hence, = 1.2157.

Again consider,

T =

Differentiate above equation with respect to “V” then

=

= (1)

Substitute P value in above equation we get

=

= 29.1841

Hence, = 29.1841.

The estimate of uncertainty is given by

=

=

= 0.0688.

Hence, = 0.0688.

Therefore, The estimate of uncertainty is T =  294.84710.0688 kelvin.

Step3 of 4

b).

We have P = 242.52, T = 290.11,

Consider,

T =

V =

=

=

= 9.9406

Hence, V = 9.9406.

Now,

V =

Differentiate above equation with respect to “P” then

= (

= (8.31)T

= (8.31)T()

= (8.31)T(-)

= -(8.31)T()

= -(8.31)(290.11)(

=  -0.0409

Hence, = -0.0409.

Consider,

V =

Differentiate above equation with respect to “T” then

= (

= ()

= ()(1)

= ()

= 0.03426

Hence, = 0.0324.

The estimate of uncertainty is given by

=

=

= 0.001421.

Hence, = 0.001421.

Therefore, The estimate of uncertainty is V =  9.94060.001421 liters.

Step4 of 4:

c).

We have V = 10.103,  T = 290.11, .

Consider,

T =

(8.31)T = PV

P =

=

=

= 238.6235

Step 2 of 3

Step 3 of 3

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