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vaieri [72.5K]
2 years ago
8

1.00 mole of an ideal monoatomic gas at STP first undergoes isothermal expansion so that the volume at b is 2.5 times the volume

at a. Next, heat is extracted at a constant volume so that the pressure drops. The gas is then compressed adiabatically back to the original state. Calculate the pressure at c.
Chemistry
1 answer:
MrRa [10]2 years ago
8 0

Answer:

the pressure at c = 0.27 atm

Explanation:

Given that:

number of moles (n) = 1.0 moles

Value of gamma in the monoatomic gas (γ) = 5/3

During an isothermal expansion, the volume at b is = 2.5 times the volume at a ; this implies that:

V_b = 2.5 V_a

∴ To calculate  the pressure at c from a; the process is adiabatic compression; so we apply:

P_aV_a^\gamma=P_cV_c^\gamma

\frac{P_c}{P_a}=[\frac{V_a}{V_c}]^{(2/3)

\frac{P_c}{1.0 atm}=[\frac{1}{2.5}]^{(2/3)

P_c=0.27 atm

Thus, the pressure at c = 0.27 atm

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5 0
3 years ago
a sample of 3.00 g of so2 (g)originally in a 5.00 L vesselat 21 degee Celsius is transferred to a 10.0 L vessel at 26 degree Cel
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Answer:

1) The partial pressure of SO₂ gas in the larger container = 0.115 atm.

2) The partial pressure of N₂ gas in the larger container = 0.206 atm.

3) The total pressure in the vessel = 0.321 atm.

Explanation:

  • To calculate the partial pressure of each gas, we can use the general law of ideal gas: PV = nRT.

where, P is the partial pressure of the gas in atm,

V is the volume of the vessel in L,

n is the no. of moles of the gas,

R is the general gas constant (R = 0.082 L.atm/mol.K),

T is the temperature of the gas in K.

<u><em>1) What is the partial pressure of SO₂ gas in the larger container?</em></u>

<em>∵ P = nRT/V.</em>

n = mass/molar mass = (3.0 g)/(64.066 g/mol) = 0.047 mol.

R = 0.082 L.atm/mol.K.

T = 26 °C + 273.15 = 299.15 K.

V = 10.0 L. (The volume of the new container)

∴ P = nRT/V = (0.047 mol)(0.082 L.atm/mol.K)(299.15 K)/(10.0 L) = 0.115 atm.

<u><em>2) What is the partial pressure of N₂ gas in the larger container?</em></u>

<em>∵ P = nRT/V.</em>

n = mass/molar mass = (2.35 g)/(28.0 g/mol) = 0.084 mol.

R = 0.082 L.atm/mol.K.

T = 26 °C + 273.15 = 299.15 K.

V = 10.0 L. (The volume of the new container)

∴ P = nRT/V = (0.084 mol)(0.082 L.atm/mol.K)(299.15 K)/(10.0 L) = 0.206 atm.

<u><em>3) What is the total pressure in the vessel?</em></u>

  • According to Dalton's law the total pressure exerted is equal to the sum of the partial pressures of the individual gases.

<em>∵ The total pressure in the vessel = the partial pressure of SO₂ + the partial pressure of N₂.</em>

∴ The total pressure in the vessel = 0.115 + 0.206 = 0.321 atm.

5 0
2 years ago
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