To find the pressure of the nitrogen gas, we can use the ideal gas law equation:

$$ PV = nRT $$

where:
- $ P $ is the pressure,
- $ V $ is the volume,
- $ n $ is the number of moles,
- $ R $ is the universal gas constant,
- $ T $ is the temperature in Kelvin.

Given:
- $ n = 5.0 $ mol,
- $ V = 2.0 $ L,
- $ T = 268 $ K,
- $ R = 0.0821 $ L·atm/mol·K.

We need to solve for $ P $:

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

Substitute the given values into the equation:

$$ P = \frac{(5.0 \text{ mol}) (0.0821 \text{ L·atm/mol·K}) (268 \text{ K})}{2.0 \text{ L}} $$

Calculate the numerator:

$$ (5.0) (0.0821) (268) = 110.114 $$

Now, divide by the volume:

$$ P = \frac{110.114}{2.0} = 55.057 \text{ atm} $$

Rounding to two significant figures, the pressure is approximately:

$$ P \approx 55 \text{ atm} $$

Therefore, the correct answer is:

C. 55 atm

Question

To find the pressure of the nitrogen gas, we can use the ideal gas law equation:

$$ PV = nRT $$

where:
- $ P $ is the pressure,
- $ V $ is the volume,
- $ n $ is the number of moles,
- $ R $ is the universal gas constant,
- $ T $ is the temperature in Kelvin.

Given:
- $ n = 5.0 $ mol,
- $ V = 2.0 $ L,
- $ T = 268 $ K,
- $ R = 0.0821 $ L·atm/mol·K.

We need to solve for $ P $:

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

Substitute the given values into the equation:

$$ P = \frac{(5.0 \text{ mol}) (0.0821 \text{ L·atm/mol·K}) (268 \text{ K})}{2.0 \text{ L}} $$

Calculate the numerator:

$$ (5.0) (0.0821) (268) = 110.114 $$

Now, divide by the volume:

$$ P = \frac{110.114}{2.0} = 55.057 \text{ atm} $$

Rounding to two significant figures, the pressure is approximately:

$$ P \approx 55 \text{ atm} $$

Therefore, the correct answer is:

C. 55 atm

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