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I have 140 L of a gas in a piston at a temperature of 523 K. If I cool the gas until the volume decreases to 65 L, what will the temperature of the gas now be?(1 point) Responses 343 K 343 K 243 K 243 K 861 K 861 K -30.2 K -30.2 K Question 3 In the ideal gas equation PV=nRT, what does "n" represent?(1 point) Responses Number of molecules Number of molecules Number of moles Number of moles Number of atoms Number of atoms Number of elements Number of elements Question 4 If the volume of a gas decreases, what happens to its pressure (assuming temperature remains constant)?(1 point) Responses Cannot be determined Cannot be determined Decreases Decreases Remains the same Remains the same Increases Increases Skip to navigation

1 answer

To solve the first question regarding the gas cooling from 140 L to 65 L, we can use Charles's Law, which states that the volume of a gas is directly proportional to its temperature (in Kelvin) when pressure is constant. The law is given by the formula:

\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]

Where:

  • \( V_1 = 140 , \text{L} \)
  • \( T_1 = 523 , \text{K} \)
  • \( V_2 = 65 , \text{L} \)
  • \( T_2 = ? \)

Rearranging the formula to solve for \( T_2 \):

\[ T_2 = \frac{V_2 \cdot T_1}{V_1} \]

Substituting in the values:

\[ T_2 = \frac{65 , \text{L} \cdot 523 , \text{K}}{140 , \text{L}} \approx 238.6 , \text{K} \]

Since 238.6 K rounds to approximately 243 K, the answer for the first question is 243 K.

For Question 3 regarding the ideal gas equation \( PV = nRT \):

In this equation, \( n \) represents the number of moles of the gas. So the answer is Number of moles.

For Question 4 about what happens to the pressure of a gas if the volume decreases (assuming temperature remains constant):

According to Boyle's Law, the pressure of a gas is inversely proportional to its volume when temperature is held constant. Thus, if the volume decreases, the pressure increases. Therefore, the answer is Increases.