Question

A steel bottle contains argon gas at STP. What is the final pressure if the temperature is changed to 115°C?

Answer 1

Using the Combined Gas Law and keeping volume constant, the final pressure of argon gas when the temperature is increased to 115°C, assuming STP conditions initially, is 1.42 atm.

Step-by-step explanation:

To find the final pressure of argon gas when the temperature is changed to 115°C, we use the Combined Gas Law which is derived from the Ideal Gas Law. The Combined Gas Law is P1V1/T1 = P2V2/T2, where P, V, and T refer to the pressure, volume, and temperature of the gas (with T measured in Kelvin). At STP (Standard Temperature and Pressure), we know that the pressure is 1 atm and the temperature is 0°C or 273.15 K. If only the temperature changes and the volume remains constant, we can simplify the equation to P1/T1 = P2/T2. We can solve for the final pressure P2 using the initial conditions and the final temperature (which is 115°C or 388.15 K).

To calculate the final pressure, the formula becomes:

P2 = (P1 × T2) / T1

P2 = (1 atm × 388.15 K) / 273.15 K

P2 = 1.42 atm

Therefore, the final pressure of the argon gas at 115°C is 1.42 atm.

Answer 2

Final pressure is 1.42atm

Step-by-step explanation:

Based on Gay-Lussac's law, pressure of a gas is directely proportional to its absolute temperature. The equation of this law is:

P₁T₂ = P₂T₁

Where P is pressure and T is absolute temperature of 1, initial state and 2, final state of the gas.

In the problem, initial conditions are Standard Temperature and Pressure, STP, that are 1 atm and 273.15K.

If the final temperature is 115°C = 388.15K (115°C + 273.15 = 388.15K), using Gay-Lussac's law:

P₁T₂ = P₂T₁

1atmₓ388.15K = P₂ₓ273.15K

1.42atm = P₂

Final pressure is 1.42atm