calculate the parameter which is missing in the following table. R = 0.082

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Using the ideal gas law, calculate the parameter which is missing in the following table. R = 0.082 (atm x L)/(Mole K) or 62.4 (mmHg L)/(mole K)  Pressure Volume # of moles Temperature 6.00 atm 350. mL moles 400. �C 900. mm Hg mL 0.400 moles 400. K 3.50 atm 250. L 10.0 moles _�C _. mm Hg 44.8 L 2.00 mole 273 K

  1. Using the ideal gas law, calculate the parameter which is missing in the following table. R = 0.082 (atm x L)/(Mole K) or 62.4 (mmHg L)/(mole K)ÂÂ

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Pressure Volume # of moles Temperature
6.00 atm 350. mL _ moles 400. �C

  1. mm Hg mL 0.400 moles 400. K
    3.50 atm 250. L 10.0 moles __    �C
    _. mm Hg 44.8 L 2.00 mole 273 K

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The Correct Answer and Explanation is:

Let’s use the Ideal Gas Law: PV=nRTPV = nRT

Where:

  • PP = Pressure
  • VV = Volume
  • nn = Number of moles
  • RR = Ideal Gas Constant
  • TT = Temperature (in Kelvin)

We’ll solve each missing value step-by-step.

1. Missing Moles

Given:

  • Pressure P=6.00P = 6.00 atm
  • Volume V=350.V = 350. mL = 0.350 L
  • Temperature ( T = 400^\circ C = 673 K**
  • R=0.082 L\cdotpatmmol\cdotpKR = 0.082 \, \frac{\text{L·atm}}{\text{mol·K}}

Using: n=PVRT=6.00×0.3500.082×673=2.155.186≈0.0381 moln = \frac{PV}{RT} = \frac{6.00 \times 0.350}{0.082 \times 673} = \frac{2.1}{55.186} \approx \boxed{0.0381 \text{ mol}}

2. Missing Volume

Given:

  • Pressure = 900 mmHg
  • Moles = 0.400 mol
  • Temperature = 400 K
  • R=62.4 mmHg\cdotpLmol\cdotpKR = 62.4 \, \frac{\text{mmHg·L}}{\text{mol·K}}

V=nRTP=0.400×62.4×400900=9984900≈11.1 LV = \frac{nRT}{P} = \frac{0.400 \times 62.4 \times 400}{900} = \frac{9984}{900} \approx \boxed{11.1 \text{ L}}

3. Missing Temperature (°C)

Given:

  • Pressure = 3.50 atm
  • Volume = 250 L
  • Moles = 10.0 mol
  • R=0.082R = 0.082

T=PVnR=3.50×25010.0×0.082=8750.82≈1067.07K⇒T∘C=1067.07−273=794∘CT = \frac{PV}{nR} = \frac{3.50 \times 250}{10.0 \times 0.082} = \frac{875}{0.82} \approx 1067.07 K \Rightarrow T_{^\circ C} = 1067.07 – 273 = \boxed{794^\circ C}

4. Missing Pressure (mmHg)

Given:

  • Volume = 44.8 L
  • Moles = 2.00 mol
  • Temperature = 273 K
  • R=62.4R = 62.4

P=nRTV=2.00×62.4×27344.8=34049.644.8≈760 mmHgP = \frac{nRT}{V} = \frac{2.00 \times 62.4 \times 273}{44.8} = \frac{34049.6}{44.8} \approx \boxed{760 \text{ mmHg}}

Summary Table:

PressureVolumeMolesTemperature
6.00 atm350. mL0.0381 mol400°C (673 K)
900 mm Hg11.1 L0.400 mol400 K
3.50 atm250. L10.0 mol794°C (1067 K)
760 mm Hg44.8 L2.00 mol273 K

The ideal gas law is an equation that describes the relationship between pressure (P), volume (V), temperature (T), and the number of moles (n) of a gas. It’s represented as PV=nRTPV = nRT, where RR is the universal gas constant. The gas constant value depends on the units used: 0.082 L·atm/mol·K when pressure is in atm and 62.4 L·mmHg/mol·K when pressure is in mmHg.

To solve for a missing variable, the equation can be rearranged. For example, to find moles nn, we use n=PVRTn = \frac{PV}{RT}. Temperature must always be in Kelvin, which is obtained by adding 273 to the Celsius temperature. Volume should be in liters (L).

In the first case, converting 350 mL to liters and Celsius to Kelvin allows us to calculate the number of moles. In the second case, we rearranged the formula to solve for volume. The third case needed solving for temperature in Kelvin, then converted to Celsius. The final case involved solving for pressure using mmHg units.

Each value was computed by substituting the known variables into the formula and solving algebraically. These calculations reinforce the reliability of the ideal gas law in predicting the behavior of gases under varying conditions, assuming ideal behavior.

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