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A large cylindrical tank contains 0.750 m3 of nitrogen gas at 27°C and 7.50 * 103 Pa (absolute pressure). The tank has a tight‐fitting piston that allows the volume to be changed. What will be the pressure if the volume is decreased to 0.480 m3 and the temperature is increased to 157°C?

User Debajit
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2 Answers

3 votes

Final answer:

To find the final pressure in the tank, use the ideal gas law equation PV = nRT. Convert the temperatures from Celsius to Kelvin and calculate the initial number of moles. Plug in the values for the final volume, initial moles, ideal gas constant, and final temperature to solve for the final pressure.

Step-by-step explanation:

To determine the final pressure in the tank, we can use the ideal gas law equation, which states: 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.

First, we need to convert the given temperatures from Celsius to Kelvin. The initial temperature is 27°C + 273 = 300K, and the final temperature is 157°C + 273 = 430K.

Next, we can calculate the initial number of moles of nitrogen gas using the ideal gas law equation for the initial conditions: (7.50 * 10^3 Pa) * (0.750 m^3) = n * (8.31 J/(mol * K)) * (300K). Solving for n gives us an initial number of moles of approximately 0.246 moles.

Using the ideal gas law equation for the final conditions, we can solve for the final pressure. (Pf) * (0.480 m^3) = (0.246 moles) * (8.31 J/(mol * K)) * (430K). Solving for Pf gives us a final pressure of approximately 10,104 Pa.

User Tony Isaac
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6 votes

Answer: The answer is 68142.4 Pa

Explanation:

Given that the initial properties of the cylindrical tank are :

Volume V1= 0.750m3

Temperature T1= 27C

Pressure P1 =7.5*10^3 Pa= 7500Pa

Final properties of the tank after decrease in volume and increase in temperature :

Volume V2 =0.480m3

Temperature T2 = 157C

Pressure P2 =?

Applying the gas law equation (Charles and Boyle's laws combined)

P1V1/T1 = P2V2/T2

(7500 * 0.750)/27 =( P2 * 0.480)/157

P2 =(7500 * 0.750* 157) / (0.480 *27)

P2 = 883125/12.96

P2 = 68142.4Pa

Therefore the pressure of the cylindrical tank after decrease in volume and increase in temperature is 68142.4Pa

User McMath
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