Question

a gas has an initial volume of 212 cm³ at a temperature of 293 k and a pressure of 0.98 atm. what is the final pressure of the gas if the volume decreases to 196 cm³ and the temperature of the gas increases to 308 k?
0.88 atm
0.95 atm
1.0 atm
1.1 atm

Explanation:

Step1: Recall Combined Gas Law

The combined gas law is \(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\), where \(P_1 = 0.98\) atm, \(V_1=212\) \(cm^3\), \(T_1 = 293\) K, \(V_2 = 196\) \(cm^3\), \(T_2=308\) K, and we need to find \(P_2\).

Step2: Rearrange the formula to solve for \(P_2\)

From \(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\), we get \(P_2=\frac{P_1V_1T_2}{T_1V_2}\).

Step3: Substitute the values into the formula

Substitute \(P_1 = 0.98\) atm, \(V_1 = 212\) \(cm^3\), \(T_1=293\) K, \(V_2 = 196\) \(cm^3\), \(T_2 = 308\) K into the formula:

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Answer:

1.1 atm (assuming the last option is 1.1 atm, as the text is a bit unclear but based on calculation the final pressure is approximately 1.1 atm)