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Keith_Richards [23]
3 years ago
11

A sample of ammonia gas at 298 K has a volume of 3.5 L at a pressure of 170.2 kPa, what is the new pressure if the gas is compre

ssed to a volume of 1.35 L at 293 K?
Chemistry
2 answers:
Maru [420]3 years ago
5 0

Answer:

430 kPa

Explanation:

Use the formula PV/T = PV/T where the left side of the equation refers to the initial values and the right side refers to the final values: (170.2 kPa)(3.5 L)/(298 K) = (P)(1.35 L)/(293 K). The P represents the unknown pressure. Solving this equation gives us a value for P of 433.855 kPa, but because of significant figures (the two from 3.5 L), we round to 430 kPa.

yarga [219]3 years ago
3 0

Answer:

The final pressure of the gas comes out to be 433.85 KPa

Explanation:

Initial volume of gas = V = 3.5 L

Initial pressure of gas =  P = 170.2 KPa

Initial temperature = T = 298 K

Final volume of gas = V' = 1.35 L

Final temperature of gas = T' = 293 K

Assuming the number of moles of gas remains constant.

The final pressure of the gas is obtained by the following equation shown below

\displaystyle \frac{PV}{T} = \displaystyle \frac{P'V'}{T'} \\\displaystyle \frac{170.2 \textrm{ KPa}\times 3.5 \textrm{ L}}{298\textrm{ K}} = \displaystyle \frac{P'\times 1.35 \textrm{ L}}{293 \textrm{ K}} \\P' = 433.85 \textrm{ KPa}

Final pressure of gas = 433.85 KPa

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7 0
3 years ago
Read 2 more answers
The molar heat capacity of ethane is represented in the temperature range 298 K to 400 K by the empirical expression Cp,m in J K
BabaBlast [244]

Answer:

-88.66 kJ/mol

Explanation:

The expressions of heat capacity (Cp,m) for C(s) and for H₂(g) are:

C(s):  Cp,m/(J K-1 mol-1) = 16.86 + (4.77T/10³) - (8.54x10⁵/T²)

H₂(g): Cp,m/(J K-1 mol-1) = 27.28 + (3.26T/10³) + (0.50x10⁵/T²)

Cp = A + BT + CT⁻²

For the Kirchoff's Law:

ΔHf = ΔH°f + \int\limits^{T2}_{T1} {DCp(T)} \, dT

Where ΔH°f is the enthalpy at 298 K, T1 is 298 K, T2 is the temperature given (373 K), and DCp is the variation of Cp (products less reactants). ΔH°f  for ethene is -84.68 kJ/mol and the reaction is:

2C(s) + 3H₂(g) → C₂H₆

So, DCp:

dA = A(C₂H₆) - [2xA(C) + 3xA(H₂)] = 14.73 - [2x16.86 + 3x27.28] = -100.83

dB = B(C₂H₆) - [2xB(C) + 3xB(H₂)] = 0.1272 - [2x4.77x10⁻³ + 3x3.26x10⁻³] = 0.10788

dC = C(C₂H₆) - [2xC(C) + 3xC(H₂)] = 0 - (2x(-8.54x10⁵) + 3x0.50x10⁵) = 15.58x10⁵

dCp = -100.83 + 0.10788T + 15.58x10⁵T⁻²

\int\limits^{373}_{298} {-100.83 + 0.10788T + 15.58x10^5T^{-2}} \, dT = -3796.48 J/mol = -3.80 kJ/mol (solved by a graphic calculator)

ΔHf = -84.68 - 3.80

ΔHf = -88.66 kJ/mol

7 0
3 years ago
If 5.40 kcal of heat is added to 1.00 kg of water at 100⁰C, how much steam at 100⁰C is produced? Show all calculations leading t
Eddi Din [679]
Heat of vaporization of water = 540 cal/g
m=1000 g
energy supplied = 5.4 kcal=5400 cal
T1=T2=100 deg. C
so all energy is used to convert to steam
mass of steam produced = 5400 cal /(540 cal/g)
=5400/540 cal *g/cal
=10 g
8 0
3 years ago
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viktelen [127]

Answer:

conduction

Explanation:

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