3. One mole of oxygen gas is at a pressure of 5.30 atm and a temperature of 25.0

Question

3. One mole of oxygen gas is at a pressure of 5.30 atm and a temperature of 25.0°C.
(a) If the gas is heated at constant volume until the pressure triples, what is the final temperature?
°C
(b) If the gas is heated so that both the pressure and volume are doubled, what is the final temperature?
°C
4. A cylinder with a movable piston contains gas at a temperature of 28.0°C, a volume of 1.90 m3, and an absolute pressure of 0.200 multiplied by 105 Pa. What will be its final temperature if the gas is compressed to 0.900 m3 and the absolute pressure increases to 0.800 multiplied by 105 Pa?
°C

Explanation / Answer

Given Data 3. Number of moles , n = 1     Initial Pressure, P1 = 5.30 atm     Initial temperature, T1 = 250C ------------------------------------------------------------------------------ (a) Since, Volume is constant , V1 = V2     Final pressure , P2 = 3 P1    Using the ideal gas equation,                          PV = n R T                  (P1 / P2) = (T1 / T2 )                (P1 / 3P1) = (T1 / T2 )                           T2 = 3 T1                                 = 3( 250C)                                 = 750C               -------------------------------------------------------------------------------------- (b) If P2 = 2 P1 , and   V2 = 2V1 . Then,             P1 V1 / T1 = P2 V2 / T2            P1 V1 / T1 = (2P1)(2 V1) /  T2                        T2 = 4 T1                              = 4( 250C)                              = 1000C               --------------------------------------------------------------------------------------- 4.Initial temperature, T1 = 280C    Initial Volume, V1 = 1.90 m3    Initial pressure, P1 = 0.2 * 105 Pa    Final Volume, V2 = 0.90 m3    Final pressure, P2 = 0.8 *105 Pa    Using the ideal gas equation,                                                       PV = n R T                                            P1 V1 / T1 = P2 V2 / T2             ( 0.2 * 105 Pa)( 1.90 m3) /(280C) = (0.8 *105 Pa) ( 0.90 m3) / T2                                                         T2 = 53.050 C                        T2 = 4 T1                              = 4( 250C)                              = 1000C               --------------------------------------------------------------------------------------- 4.Initial temperature, T1 = 280C    Initial Volume, V1 = 1.90 m3    Initial pressure, P1 = 0.2 * 105 Pa    Final Volume, V2 = 0.90 m3    Final pressure, P2 = 0.8 *105 Pa    Using the ideal gas equation,                                                       PV = n R T                                            P1 V1 / T1 = P2 V2 / T2             ( 0.2 * 105 Pa)( 1.90 m3) /(280C) = (0.8 *105 Pa) ( 0.90 m3) / T2                                                         T2 = 53.050 C    Final pressure, P2 = 0.8 *105 Pa    Using the ideal gas equation,                                                       PV = n R T                                            P1 V1 / T1 = P2 V2 / T2             ( 0.2 * 105 Pa)( 1.90 m3) /(280C) = (0.8 *105 Pa) ( 0.90 m3) / T2                                                         T2 = 53.050 C                                                         T2 = 53.050 C

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