## IIT JEE Main 2017 Physics Question 11 SetA:

IIT JEE Main 2017 Physics Question 11 SetA states that the temperature of an open room of volume 30 m^{3} increases from 17 deg C to 27 deg C due to the sunshine. The atmospheric pressure in the room remains 1 x 10^{5} Pa. If n_{i} and n_{f} are the number of molecules in the room before and after heating, then n_{f} – n_{i} will be:

(1) -1.61 x 10^{23} |

(2) 1.38 x 10^{23} |

(3) 2.5 x 10^{25} |

(4) -2.5 x 10^{25} |

#### Solution

According to the problem, there is constant atmospheric pressure in the room and:

n_{1} = initial number of moles

=> n_{1} = (P_{1}V_{1})/(RT_{1})

=> n_{1} = (10^{5} X 30)/(8.3 X 290)

=> n_{1} = 1.24 X 10^{3}

n_{2} = final number of moles

=> n_{2} = (P_{2}V_{2})/(RT_{2})

=> n_{2} = (10^{5} X 30)/(8.3 X 300)

=> n_{2} = 1.20 X 10^{3}

Change in number of molecules :

n_{f} – n_{i} = (n_{2} – n_{1}) x 6.023 x 10^{23}

n_{f} – n_{i} = -2.5 X 10^{25}

#### Answer to IIT JEE Main 2017 Physics Question 11 SetA

Correct answer to given problem No 11 of IIT JEE Main Physics paper of 2017 Set A is option number (4) that is the Change in number of molecules is n_{f} – n_{i} = -2.5 X 10^{25}