Question

75 mL of a gas is at 690 mmHg at -45°C. What will be the new volume at
STP? *

Answer 1

56.9 mL

Step-by-step explanation:

To solve this problem, we can use the equation of state for an ideal gas, which states that:

`pV=nRT`

where

p is the pressure of the gas

V is its volume

n is the number of moles in the gas

R is the gas constant

T is the absolute temperature of the gas

For a gas undergoing a transformation, n and R remain constant, so we can rewrite the equation as:

`(p_1 V_1)/(T_1)=(p_2 V_2)/(T_2)`

where in this case:

`p_1 = 690 mmHg = 0.908 atm` is the initial pressure of the gas

`V_1=75 mL` is the initial volume

`T_1=-45^(\circ)+273=228 K` is the initial temperature of the gas

`p_2 = 1 atm` is the final pressure (at STP)

`T_2 = 273 K` is the final temperature (at STP)

Solving for V2, we find the final volume of the gas:

`V_2=(p_1 V_1T_2)/(T_1p_2)=((0.908)(75)(228))/((273)(1))=56.9 mL`