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A sample of gas has a volume of 4.20 Lat 333 K and 1 atm of pressure. What pressure in atmospheres (atm) would be required for the gas to expand to 5.00 Lif the temperature cooled in the process to 300 K

User Jwiklund
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Answer:

The pressure required for the gas to expand to 5.00 L if the temperature cooled in the process to 300 K is 0.84 atm.

Step-by-step explanation:

To solve this problem, we can use the ideal gas law:

PV = nRT

where:

P is the pressure in atm

V is the volume in L

n is the number of moles of gas

R is the gas constant (0.08206 L atm / mol K)

T is the temperature in K

Since we are dealing with the same sample of gas, the number of moles of gas remains constant. Therefore, we can write the ideal gas law as follows:

P1V1 = P2V2

where:

P1 is the initial pressure (1 atm)

V1 is the initial volume (4.20 L)

P2 is the final pressure (unknown)

V2 is the final volume (5.00 L)

Substituting the given values into the equation, we get:

1 atm * 4.20 L = P2 * 5.00 L

P2 = 0.84 atm

Therefore, the pressure required for the gas to expand to 5.00 L if the temperature cooled in the process to 300 K is 0.84 atm.

User Mariz
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