The volume V of a fixed amount of a gas varies directly as the temperature T and inversely as the pressure P. Suppose that V=120 cm" when T=260 kelvin kg Find the pressure when T=380 kelvin and V= 152 cm cm and P= 13 kg cm
Unitary Method
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Speed, Time, and Distance
Imagine you and 3 of your friends are planning to go to the playground at 6 in the evening. Your house is one mile away from the playground and one of your friends named Jim must start at 5 pm to reach the playground by walk. The other two friends are 3 miles away.
Profit and Loss
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Units and Measurements
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![### Problem Statement:
The volume \( V \) of a fixed amount of a gas varies directly as the temperature \( T \) and inversely as the pressure \( P \). Suppose that \( V = 120 \, \text{cm}^3 \) when \( T = 260 \, \text{kelvin} \) and \( P = 13 \, \frac{\text{kg}}{\text{cm}^2} \). Find the pressure when \( T = 380 \, \text{kelvin} \) and \( V = 152 \, \text{cm}^3 \).
### Explanation:
The direct and inverse proportionality can be expressed mathematically using the equation:
\[ V = \frac{kT}{P} \]
where \( k \) is the proportionality constant.
Given:
- \( V = 120 \, \text{cm}^3 \)
- \( T = 260 \, \text{kelvin} \)
- \( P = 13 \, \frac{\text{kg}}{\text{cm}^2} \)
First, we need to find the proportionality constant \( k \):
\[ 120 = \frac{k \cdot 260}{13} \]
Calculating \( k \):
\[ k = \frac{120 \cdot 13}{260} \]
\[ k = 6 \]
Now, we need to find the new pressure \( P \) when:
- \( T = 380 \, \text{kelvin} \)
- \( V = 152 \, \text{cm}^3 \)
Using the same equation with the known \( k \):
\[ V = \frac{6 \cdot T}{P} \]
\[ 152 = \frac{6 \cdot 380}{P} \]
Solving for \( P \):
\[ 152P = 6 \cdot 380 \]
\[ 152P = 2280 \]
\[ P = \frac{2280}{152} \]
\[ P = 15 \, \frac{\text{kg}}{\text{cm}^2} \]
### Answer:
The pressure when \( T = 380 \, \text{kelvin} \) and \( V = 152 \, \text{cm}^3 \) is \( 15 \, \frac](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F291d7961-38c8-46f7-a661-8bb6ad5e5399%2F4c258c5e-13bc-42e1-93a0-24ad99c7a075%2Fpzglo7k_processed.jpeg&w=3840&q=75)
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