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raketka [301]
3 years ago
8

A balloon containing helium gas expands from 230

Chemistry
1 answer:
andre [41]3 years ago
5 0

Answer:

1.0 *10^(-4) mol

Explanation:

For gases:

n1/n2 = V1/V2

n1/3.8*10^(-4) mol = 230 mL/ 860 mL

n1 = 3.8*10^(-4)*230/860 = 1.0 *10^(-4) mol

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55 L of a gas at 25oC has its temperature increased to 35oC. What is its new volume?
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Answer:

Approximately 56.8 liters.

Assumption: this gas is an ideal gas, and this change in temperature is an isobaric process.

Explanation:

Assume that the gas here acts like an ideal gas. Assume that this process is isobaric (in other words, pressure on the gas stays the same.) By Charles's Law, the volume of an ideal gas is proportional to its absolute temperature when its pressure is constant. In other words

\displaystyle V_2 = V_1\cdot \frac{T_2}{T_1},

where

  • V_2 is the final volume,
  • V_1 is the initial volume,
  • T_2 is the final temperature in degrees Kelvins.
  • T_1 is the initial temperature in degrees Kelvins.

Convert the temperatures to degrees Kelvins:

T_1 = \rm 25^{\circ}C = (25 + 273.15)\; K = 298.15\; K.

T_2 = \rm 35^{\circ}C = (35 + 273.15)\; K = 308.15\; K.

Apply Charles's Law to find the new volume of this gas:

\displaystyle V_2 = V_1\cdot \frac{T_2}{T_1} = \rm 55\;L \times \frac{308.15\; K}{298.15\; K} = 56.8\; L.

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