A region of the galaxy where new stars are forming-contains a very tenuous gas with 100 #a t oms ##/cm^3#. This gas is heated to #7500 K# by ultraviolet radiation from nearby stars. What is the Gas Pressure in ATM?
The pressure of the gas under these conditions would be
An accurate estimate is provided by the ideal gas law at this extremely low pressure.
wherein
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To find the gas pressure in ATM, you can use the ideal gas law equation:
[PV = nRT]
Where:
- (P) is the pressure in atmospheres (ATM)
- (V) is the volume of the gas in cubic centimeters (cm³)
- (n) is the number of moles of gas
- (R) is the ideal gas constant ((0.0821 \frac{L \cdot atm}{K \cdot mol}))
- (T) is the temperature in Kelvin (K)
Given:
- Gas density ((n/V)) = (100) atoms/cm³
- Temperature ((T)) = (7500) K
We need to find the pressure ((P)). First, we need to calculate the number of moles of gas using the gas density and volume. Then, we can rearrange the ideal gas law equation to solve for pressure ((P)).
[n = \frac{100}{6.022 \times 10^{23}} \times V]
[P = \frac{nRT}{V}]
Substitute the values and solve for (P):
[P = \frac{\left(\frac{10
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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