Initial Conditions | Final Conditions (s.t.p.) |
---|---|

P_{1} = 700 mm of Hg |
P_{2} = 760 mm of Hg |

V_{1} = 10 lit |
V_{2} = x lit |

T_{1} = 27 + 273 K |
T_{2} = 273 K |

Using the gas equation,

⇒ P_1V_1\over T_1=P_2V_2\over T_2

Substituting the values we get,

⇒ 700 × 10\over 300=760× x\over 273

⇒ x = 700 × 10 × 273\over 300 × 760

⇒ x = 1911\over 228 = 8.38 lit

∴ 70 g of the gas occupies 22.4 lit. at s.t.p.

If $y$ g of the gas occupies 8.38 lits, then

y = 70\over 22.4 ×8.38

= 26.18

Hence, **weight of substance is 26.18 g**