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

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