CHEMISTRY HIGH SCHOOL

A rigid cylinder contains a sample of gas at STP. What is the pressure of this gas after the sample is heated to 410 K?(1) 1.0 atm (3) 0.67 atm
(2) 0.50 atm (4) 1.5 atm

Answers

Answer 1
Answer:

Answer:(4) 1.5 atm

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

(P_1V_1)/(T_1)=(P_2V_2)/(T_2)

where,

P_1 = initial pressure of gas at STP= 1 atm

P_2 = final pressure of gas = ?

V_1 = initial volume of gas at STP = V L

V_2 = final volume of gas = V L

T_1 = initial temperature of gas at STP= 273K

T_2 = final temperature of gas = 410 K

Now put all the given values in the above equation, we get the final pressure of gas.

(1atm* VL)/(273K)=(P_2* VL)/(410K)

P_2=1.50atm

Therefore, the pressure of gas will be 1.50 atm.
Answer 2
Answer: To solve the problem, we assume the sample to be ideal. Then, we use the ideal gas equation which is expressed as PV = nRT. From the conditions of the gas sample, we calculate the pressure as follows: (NOTE: At STP 1 mol is equal to 22.4 L of a molecule)

P = nRT / V
P = (1 mol) (8.314 Pa m^3/ mol K) (298.15 K) / .0224 L
P = 110660 Pa or 1.0 atm

Thus, the answer is the value in option 1.

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Answers

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Answer:

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Explanation:

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Answers

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Learn more about nuclear reactions here: brainly.com/question/14019172

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Explanation:

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