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2 years ago

A gas has a volume of 350 mL at 45 oK. If the volume changes to 400 mL, what is the new temperature?

Chemistry

1 answer:

elena-14-01-66 [18.8K]2 years ago

4 0

Answer:

T₂ = 39.4 °K

Explanation:

Because you are only dealing with volume and temperature, you can use Charles' Law to find the missing value. The formula looks like this:

V₁ / T₁ = V₂ / T₂

In this formula, "V₁" and T₁" represent the initial volume and temperature. "V₂" and "T₂" represent the final volume and temperature. You have been given values for all of the variables except for "T₂". Therefore, by plugging these values into the formula, you can simplify to find the answer.

V₁ = 350 mL T₁ = 45 °K

V₂ = 400 mL T₂ = ?

V₁ / T₁ = V₂ / T₂ <----- Charles' Law formula

(350 mL)(45 °K) = (400 mL)T₂ <----- Insert values into variables

15750 = (400 mL)T₂ <----- Multiply left side

39.4 = T₂ <----- Divide both sides by 400

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Use the unbalanced equation NH3 + O2 = NO + H2O

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2 years ago

certain gas occupies 6.24L at pressure of 760 mm Hg.If the pressure is reduced to 60.0mm Hg,what would the new volume be?

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Answer:

79.04 L

Explanation:

We are given;

Initial Volume; V1 = 6.24L

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Final pressure; P2 = 60.0mm Hg

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2 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

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Answer:

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Explanation:

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2 years ago