At what temperature, the rate of effusion of $N_{2}$ would be 1.625 times the rate of effusion of $S O_{2}$ at 500C?
A- 110K
B- 173K
C- 373K
D- 273K

Question

At what temperature, the rate of effusion of $N_{ 2 }$  would be 1.625 times the rate of effusion of $SO_{2}$ at 500C?A-   110KB-    173KC-    373KD-    273K

Vedantu Content Team · Accepted Answer

Hint: We need not calculate the exact rate of effusion as we are given with relative rates of 2 gases. Try to find out how the rate of effusion depends on the temperature of the gas. Think about the factors that affect the rate of effusion and compare them for both gases. Complete answer:We know that effusion is the movement of gas through a small hole. According to graham’s law, this rate of effusion is inversely proportional to the square root of the molecular mass of the gas. The rate of effusion is the proportional velocity of the gas. The velocity of a gas refers to the root mean square velocity. $v_{ rms }\quad =\quad \sqrt { \dfrac { 3RT }{ M }  } $  where R is the universal gas constant, T is the temperature of the gas and M is the molecular weight. Given the rate of effusion of $N_{ 2 }$ would be 1.625 times the rate of effusion of $SO_{ 2 }$.We know that rate of effusion of a gas is proportional to the velocity of gas so the ratio of rates of effusion of 2 gases is equal to the ratio of velocities of gases.  $$\dfrac { rate\quad of\quad N_{ 2 } }{ rate\quad of\quad { SO }_{ 2 } } \quad =\quad \dfrac { \sqrt { \dfrac { 3RT_{ N_{ 2 } } }{ M_{ N_{ 2 } } }  }  }{ \sqrt { \dfrac { 3RT_{ SO_{ 2 } } }{ M_{ SO_{ 2 } } }  }  }$$$$1.625\quad =\quad \sqrt { \dfrac { M_{ SO_{ 2 } }T_{ N_{ 2 } } }{ M_{ N_{ 2 } }T_{ SO_{ 2 } } }  } $$Molecular weights of nitrogen molecules and Sulphur dioxide are 28 and 64 respectively.The temperature of Sulphur dioxide is 323K.$$1.625\quad =\quad \sqrt { \dfrac { T_{ N_{ 2 } }\times 64 }{ 323\times 28 }  } $$$$\sqrt { T_{ N_{ 2 } } } \quad =\quad \sqrt { \dfrac { \left( 1.625 \right) ^{ 2 }\times 28\times 323 }{ 64 }  }$$$$T_{ N_{ 2 } }\quad =\quad \dfrac { (1.625)^{ 2 }×28×323 }{ 64 } =\quad 373.15K$$Therefore, the temperature of nitrogen gas is nearly 373K. Option C is correct.Note: Rate of the effusion is directly proportional to the square root of the temperature because as the temperature of the gas increases the velocity of the gas increases as a result rate of effusion also increases.

At what temperature, the rate of effusion of N2 would be 1.625 times the rate of effusion of SO2 at 500C?A- 110KB- 173KC- 373KD- 273K

Repeaters Course for NEET 2022 - 23

At what temperature, the rate of effusion of $N_{2}$ would be 1.625 times the rate of effusion of $S O_{2}$ at 500C?
A- 110K
B- 173K
C- 373K
D- 273K