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A cylinder with a moveable piston contains 216 mL of nitrogen gas at a pressure of 1.42 atm and a temperature of 293 K.?
What must the final volume be for the pressure of the gas to be 1.52 atm at a temperature of 332 K?
3 Answers
- JimLv 73 weeks ago
PV=nRT <<<memorize!!
n is moles, R is gas constant (use correct one!), T is temp in K (C+273.15)
Since you have a before and after with the same n &R simplifies to
PV/T initial = nR = PV/T final
Now sub in your numbers and solve
SI Units R values:
8.31446261815324 J⋅K−1⋅mol−1
8.31446261815324 m3⋅Pa⋅K−1⋅mol−1
8.31446261815324 kg⋅m2·K−1⋅mol−1s−2
8.31446261815324×103 L⋅Pa⋅K−1⋅mol−1
8.31446261815324×10−2 L⋅bar⋅K−1⋅mol−1
0.082057 L atm mol-1K-1
62.36L·torr/mol·K = L·mmHg/mol·K
- hcbiochemLv 73 weeks ago
P1V1/T1 = P2V2/T2
1.42 atm (216 mL) / 293K = 1.52 atm (V2) / 332 K
V2 = 229 mL