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What is the gauge pressure (in atm) in a 25.0°C car tire containing 3.41 mol of gas in a 28.0 L volume? (Enter your answer to at least 2 decimal places.) (b) What will its gauge pressure be (in atm) if you add 1.50 L of gas originally at atmospheric pressure and 25.0°C ? Assume the temperature returns to 25.0°C and the volume remains constant. (Enter your answer to at least 2 decimal places.)

User Xszaboj
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Final answer:

The gauge pressure in the car tire at 25.0°C with 3.41 mol of gas in a 28.0 L volume is 2.76 atm. After adding 1.50 L of gas at atmospheric pressure, the gauge pressure increases slightly to 2.87 atm.

Step-by-step explanation:

Calculating Gauge Pressure in a Car Tire

To calculate the gauge pressure in the car tire, we will use the Ideal Gas Law, which states that PV=nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the universal gas constant, and T is the temperature in Kelvin.

(a) First, we convert the temperature to Kelvin: T = 25.0 + 273.15 = 298.15 K. Then we can calculate the gauge pressure using:

P = (nRT)/V

P = (3.41 mol × 0.0821 L·atm/K·mol × 298.15 K) / 28.0 L

P = (3.41 × 0.0821 × 298.15) / 28.0 = 2.76 atm

(b) If we add 1.50 L of gas at atmospheric pressure (1 atm) without changing the volume, we have an additional number of moles n' calculated by n' = PV/RT, so:



n' = (1 atm × 1.50 L) / (0.0821 L·atm/K·mol × 298.15 K)

n' = 1.50 / (0.0821 × 298.15)

n' = 0.0618 mol

We then add this to the original number of moles and recalculate the pressure:



P' = ((3.41 + 0.0618) mol × 0.0821 L·atm/K·mol × 298.15 K) / 28.0 L

P' = (3.4718 × 0.0821 × 298.15) / 28.0 = 2.87 atm

User David John Welsh
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