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Sagot :
To solve the given problem, let's carefully analyze the information and assign our variables step by step.
1. We are told that initially, the gas exerts a pressure of [tex]\( P_1 = 151 \, \text{mm Hg} \)[/tex] in an unspecified volume.
2. We need to determine the pressure, [tex]\( P_2 \)[/tex], when the gas is transferred to a container with a specified new volume of [tex]\( 0.532 \, L \)[/tex].
3. The problem directly gives us the value of [tex]\( P_2 \)[/tex].
### Assigning the variables:
- Initial pressure: [tex]\( P_1 = 151 \, \text{mm Hg} \)[/tex]
- New pressure: [tex]\( P_2 = 166 \, \text{mm Hg} \)[/tex]
- New volume: [tex]\( 0.532 \, L \)[/tex]
### Addressing the question:
We need to find what [tex]\( P_2 \)[/tex] represents in the context given. The options provided are:
A. [tex]\( 166 \, \text{mm Hg} \)[/tex]
B. [tex]\( 0.532 \, L \)[/tex]
C. unknown
Given the information in the problem, [tex]\( P_2 \)[/tex] was stated to be the pressure in the new container.
So, the correct answer is:
A. [tex]\( 166 \, \text{mm Hg} \)[/tex]
Thus, [tex]\( P_2 = 166 \, \text{mm Hg} \)[/tex].
1. We are told that initially, the gas exerts a pressure of [tex]\( P_1 = 151 \, \text{mm Hg} \)[/tex] in an unspecified volume.
2. We need to determine the pressure, [tex]\( P_2 \)[/tex], when the gas is transferred to a container with a specified new volume of [tex]\( 0.532 \, L \)[/tex].
3. The problem directly gives us the value of [tex]\( P_2 \)[/tex].
### Assigning the variables:
- Initial pressure: [tex]\( P_1 = 151 \, \text{mm Hg} \)[/tex]
- New pressure: [tex]\( P_2 = 166 \, \text{mm Hg} \)[/tex]
- New volume: [tex]\( 0.532 \, L \)[/tex]
### Addressing the question:
We need to find what [tex]\( P_2 \)[/tex] represents in the context given. The options provided are:
A. [tex]\( 166 \, \text{mm Hg} \)[/tex]
B. [tex]\( 0.532 \, L \)[/tex]
C. unknown
Given the information in the problem, [tex]\( P_2 \)[/tex] was stated to be the pressure in the new container.
So, the correct answer is:
A. [tex]\( 166 \, \text{mm Hg} \)[/tex]
Thus, [tex]\( P_2 = 166 \, \text{mm Hg} \)[/tex].
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