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A gas that exerts a pressure of 151 mm Hg in a container with a volume of [tex]\(V_1\)[/tex] will exert a pressure of 166 mm Hg when transferred to a 0.532 L container.

Let's assign our variables:
[tex]\( P_1 = 151 \, \text{mm Hg} \)[/tex]
[tex]\( V_2 = 0.532 \, \text{L} \)[/tex]

What is [tex]\( V_1 \)[/tex]?

A. 166 mm Hg
B. 0.532 L
C. unknown



Answer :

GinnyAnswer
Let's solve the problem step-by-step:

1. Assign the given variables:
- The initial pressure of the gas, [tex]\( P_1 \)[/tex], is [tex]\( 151 \, \text{mm Hg} \)[/tex].
- The final pressure of the gas, [tex]\( P_2 \)[/tex], is [tex]\( 166 \, \text{mm Hg} \)[/tex].
- The final volume of the gas, [tex]\( V_2 \)[/tex], is [tex]\( 0.532 \, \text{L} \)[/tex].

2. Identify what is being asked:
- The problem asks for the value of [tex]\( V_2 \)[/tex].

3. State the value of [tex]\( V_2 \)[/tex]:
- From the information provided, the value of [tex]\( V_2 \)[/tex] is [tex]\( 0.532 \, \text{L} \)[/tex].

Therefore, the correct choice is:
B. [tex]\( 0.532 \, \text{L} \)[/tex]

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