What is the pressure in 2 L balloon which contains 5 moles of gas at a temperature of 250 K?
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To calculate the pressure of a gas, you can use the ideal gas law: (PV = nRT), where (P) is the pressure, (V) is the volume, (n) is the number of moles, (R) is the ideal gas constant, and (T) is the temperature in Kelvin.
Given: (V = 2) L (n = 5) moles (T = 250) K (R = 0.0821) L·atm/(mol·K)
[ P = \frac{nRT}{V} ] [ P = \frac{(5 , \text{moles})(0.0821 , \text{L·atm/(mol·K)})(250 , \text{K})}{2 , \text{L}} ]
[ P \approx 257.625 , \text{atm} ]
Therefore, the pressure in the 2 L balloon containing 5 moles of gas at 250 K is approximately 257.625 atm.
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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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