All categories

3 years ago

What is the pressure of 2.50 moles of an ideal gas if it has a volume of 50.0 liters when the temperature is 300.0 K

1 answer:

7 0

Start with the ideal gas equation, PV=nRT PV=nRT and rearrange for pressure to get p=nRTV p=nRTV . You have all the necessary variables in their proper units, so plug em' into the equation to solve for pressure in units of atmospheres.

P=(2.5 mol)(300 K)(0.08206 L atm K −1 mol −1 50.0 L =1.23 atm P=(2.5 mol)(300 K)(0.08206 L atm K−1 mol−150.0 L=1.23 atm

All that needs to be done now is converting atmospheres to mm Hg Hg .

1.23 atm∗760 mm Hg1 atm =935 mm Hg 1.23 atm∗760 mm Hg1 atm=935 mm Hg .

That value makes sense, since the original pressure in atmospheres was above 1, the pressure in mm Hg Hg will be above 760.

You might be interested in

based on the article what are some of the advantages that can be given by radio frequency and microwaves

bonufazy [111]

Answer:

higher data rates are

Explanation:

transmitted

as the bandwidth

is more

more antenna gain is possible

6 0

3 years ago

Determine Z and V for steam at 250°C and 1800 kPa by the following: (a) The truncated virial equation [Eq. (3.38)] with the foll

makvit [3.9K]

Answer:

Explanation:

Given that:

the temperature $T_1$ = 250 °C= ( 250+ 273.15 ) K = 523.15 K

Pressure = 1800 kPa

The truncated viral equation is expressed as:

$\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}$

where; B = - $152.5 \ cm^3 /mol$ C = -5800 $cm^6/mol^2$

R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹

Plugging all our values; we have

$\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

$4.138*10^{-4} \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

Multiplying through with V² ; we have

$4.138*10^4 \ V ^3 = V^2 - 152.5 V - 5800 = 0$

$4.138*10^4 \ V ^3 - V^2 + 152.5 V + 5800 = 0$

V = 2250.06 cm³ mol⁻¹

Z = $\frac{PV}{RT}$

Z = $\frac{1800*2250.06}{8.314*10^3*523.15}$

Z = 0.931

b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].

The generalized Pitzer correlation is :

$T_c = 647.1 \ K \\ \\ P_c = 22055 \ kPa \\ \\ \omega = 0.345$

$T__{\gamma}} = \frac{T}{T_c}$

$T__{\gamma}} = \frac{523.15}{647.1}$

$T__{\gamma}} = 0.808$

$P__{\gamma}} = \frac{P}{P_c}$

$P__{\gamma}} = \frac{1800}{22055}$

$P__{\gamma}} = 0.0816$

$B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}$

$B_o = 0.083 - \frac{0.422}{0.808^{1.6}}$

$B_o = 0.51$

$B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}$

$B_1 = -0.282$

The compressibility is calculated as:

$Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}$

$Z = 1+ (-0.51 +(0.345* - 0.282) ) \frac{0.0816}{0.808}$

Z = 0.9386

$V= \frac{ZRT}{P}$

$V= \frac{0.9386*8.314*10^3*523.15}{1800}$

V = 2268.01 cm³ mol⁻¹

c) From the steam tables (App. E).

At $T_1 = 523.15 \ K \ and \ P = 1800 \ k Pa$

V = 0.1249 m³/ kg

M (molecular weight) = 18.015 gm/mol

V = 0.1249 × 10³ × 18.015

V = 2250.07 cm³/mol⁻¹

R = 729.77 J/kg.K

Z = $\frac{PV}{RT}$

Z = $\frac{1800*10^3 *0.1249}{729.77*523.15}$

Z = 0.588

3 0

3 years ago

What element was named for the scientist who discovered the nucleus of the atom using gold foil

Artist 52 [7]

rutherfordium element # 104

4 0

3 years ago

What must take place for copper metal to be oxidized?

Oxana [17]

The copper would go under oxidation since it will be losing two electrons. Copper starts out with an oxidation number of zero, but in order to balance the compound of CuO with the Oxygen having an oxidation number of -2, a positive 2 is required

6 0

3 years ago

Draw the partial (valence-level) orbital diagram, and write the symbol, group number, and period number of the element:

Lapatulllka [165]

Answer:

a) The element is Manganese (Mn)

b) The element is Zirconium (Zr)

Explanation:

The step by step analysis and explanation is as shown in the attachment

4 0

3 years ago