Example Question - ideal gas law
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Calculating Gas Volume Using Charles's Law
Para resolver esta pregunta, debemos usar la ley de los gases ideales, particularmente la ley de Charles, que establece que, a presión constante, el volumen de un gas es directamente proporcional a su temperatura termodinámica. La ley de Charles se puede expresar como: \[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \] donde \( V_1 \) es el volumen inicial, \( T_1 \) es la temperatura inicial (en Kelvin), \( V_2 \) es el volumen final y \( T_2 \) es la temperatura final (en Kelvin). Debemos asegurarnos de convertir las temperaturas de grados Celsius a Kelvin sumándoles 273.15: \[ T_1 = 52°C + 273.15 = 325.15 K \] \[ T_2 = 45°C + 273.15 = 318.15 K \] Ahora podemos aplicar la ley de Charles: \[ \frac{6.5 \text{ L}}{325.15 \text{ K}} = \frac{V_2}{318.15 \text{ K}} \] Despejando para \( V_2 \) obtenemos: \[ V_2 = \frac{6.5 \text{ L} \times 318.15 \text{ K}}{325.15 \text{ K}} \] \[ V_2 = \frac{2067.975 \text{ L}\cdot\text{K}}{325.15 \text{ K}} \] \[ V_2 ≈ 6.36 \text{ L} \] Por lo tanto, el volumen que ocuparía el gas a 45°C, manteniendo la presión constante, sería aproximadamente 6.36 litros.
Ideal Gas Law and Pressure Change Relationship
The image contains a question that references the ideal gas law in the form of 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. The question asks to show the change in pressure (∆P) in terms of the change in volume (∆V) and temperature. To solve this, let's start with the given: PV = nRT Let's say the initial conditions are P, V, and T, and after the change, the conditions are P + ∆P, V + ∆V, and T. It is given that T (temperature) is constant, so n, R, and T remain unchanged. According to the ideal gas law: P * V = (P + ∆P) * (V + ∆V) Since nRT is constant, we can set both sides of the equation equal to nRT. PV = nRT (P + ∆P)(V + ∆V) = nRT Since both equations are equal to nRT: PV = (P + ∆P)(V + ∆V) Expand the right side: PV = PV + P∆V + V∆P + ∆P∆V Since PV on both sides cancel out, we're left with: 0 = P∆V + V∆P + ∆P∆V If we assume that the changes in pressure and volume are small such that ∆P∆V is negligible compared to the other terms (which is a common approximation in physics), we can simplify the equation to: 0 = P∆V + V∆P Now we can solve for ∆P in terms of ∆V: V∆P = -P∆V ∆P = -(P/V) ∆V This equation shows the change in pressure in terms of the initial pressure and volume and the change in volume, assuming constant temperature. This relation is consistent with Boyle's law, which states that for a given mass of an ideal gas at constant temperature, the product of pressure and volume is always constant.
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