The helium tank has a pressure of 5 atm at 950 °F. What will be temperature if the pressure is tripled
Answer (1)
Answer:Gay-Lussac's Law, which states that, under constant volume and amount conditions, a gas's pressure is directly proportional to its absolute temperature, can be used to solve this problem. In terms of mathematics,P₁/T₁ = P₂/T₂Where:The initial pressure is P₁.The starting absolute temperature is T₁.The last pressure is P₂.The ultimate absolute temperature is T₂. Since the gas laws demand absolute temperatures, we must first convert the starting temperature from Fahrenheit to Kelvin. The following is the formula to convert Fahrenheit (°F) to Kelvin (K):K = 5/9 (459.67 + °F)Given:P₁ = 5 atmT₁ = 950 °FConvert T₁ to K:T₁ = 5/9 (950 + 459.67)T₁ = 5/9 (1409.67)T₁ ≈ 783.15 KSince we are informed that the pressure has tripled,P₂ = 3 * P₁ = 3 * 5 atm = 15 atmThe final temperature T₂ can now be found using Gay-Lussac's Law:P₁/T₁ = P₂/T₂15 atm / T₂ = 5 atm / 783.15 KThe equation is rearranged in order to solve for T₂:T₂ = (P₂ * T₁) / P₁T₂ = (15 atm * 783.15 K) / 5 atm.T₂ = 3 * 783.15 KT₂ = 2349.45 K.The temperature in degrees Fahrenheit is requested in the question. The final temperature must therefore be converted back to Fahrenheit from Kelvin. The following formula can be used to convert Kelvin (K) to Fahrenheit (°F):°F = 9/5 (K - 273.15) + 32Fahrenheit to T₂ Conversion:°F = 9/5 (2349.45 - 273.15) + 32°F = 9/5 (2076.3) + 32 °F= 1.8 * 2076.3 + 32 °F = 3737.34 + 32°F = 3769.34 °F
