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?
0.08 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 Formula for \(P_2\)

Rearrange the combined gas law to solve for \(P_2\): \(P_2=\frac{P_1V_1T_2}{T_1V_2}\)

Step3: Substitute Values

Substitute the given values into the formula: \(P_2=\frac{0.98\times212\times308}{293\times196}\)
First, calculate the numerator: \(0.98\times212 = 207.76\); \(207.76\times308 = 207.76\times300 + 207.76\times8 = 62328+1662.08 = 63990.08\)
Then, calculate the denominator: \(293\times196 = 293\times(200 - 4)=293\times200 - 293\times4 = 58600 - 1172 = 57428\)
Now, divide the numerator by the denominator: \(P_2=\frac{63990.08}{57428}\approx1.1\) atm

Answer:

1.1 atm