Question

1. a gas is compressed from 30l at 2.0 atm and 350k to 15 l at 400k what is the new pressure?

Explanation:

Step1: Identify the gas law

This is a Boyle's Law problem, which states that for a fixed amount of gas at constant temperature, \( P_1V_1 = P_2V_2 \). Here, \( P_1 = 2.0 \, \text{atm} \), \( V_1 = 30 \, \text{L} \), \( V_2 = 15 \, \text{L} \), and we need to find \( P_2 \).

Step2: Rearrange the formula

Rearrange Boyle's Law to solve for \( P_2 \): \( P_2=\frac{P_1V_1}{V_2} \)

Step3: Substitute the values

Substitute \( P_1 = 2.0 \, \text{atm} \), \( V_1 = 30 \, \text{L} \), and \( V_2 = 15 \, \text{L} \) into the formula: \( P_2=\frac{2.0 \, \text{atm} \times 30 \, \text{L}}{15 \, \text{L}} \)

Step4: Calculate the result

Simplify the expression: \( P_2 = 4.0 \, \text{atm} \) (Wait, but the original problem might have a typo? Wait, the user's problem says "350n" and "400n" which seem like typos. Assuming the correct values are \( P_1 = 2.0 \, \text{atm} \), \( V_1 = 30 \, \text{L} \), \( V_2 = 15 \, \text{L} \) (correcting possible typos in the image). If we consider the other numbers (maybe pressure units as atm and volume as L), the calculation is as above. But if the numbers are different, let's re - check. Wait, maybe the original problem was "A gas is compressed from 30L at 2.0 atm and 350K to 15L at 400K, what is the new pressure?" Then we need to use the combined gas law \( \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} \). Let's assume that (since 350n and 400n are likely 350K and 400K). So \( P_1 = 2.0 \, \text{atm} \), \( V_1 = 30 \, \text{L} \), \( T_1 = 350 \, \text{K} \), \( V_2 = 15 \, \text{L} \), \( T_2 = 400 \, \text{K} \). Rearrange the combined gas law: \( P_2=\frac{P_1V_1T_2}{V_2T_1} \)
Substitute the values: \( P_2=\frac{2.0 \, \text{atm} \times 30 \, \text{L} \times 400 \, \text{K}}{15 \, \text{L} \times 350 \, \text{K}} \)
Calculate: \( P_2=\frac{24000}{5250}\approx4.57 \, \text{atm} \) (But since the original text has typos, we'll go with the Boyle's Law first if temperature is constant. Given the confusion in the image, but assuming the correct approach for gas law problems.)

Answer:

If using Boyle's Law (constant temperature): \( \boldsymbol{4.0 \, \text{atm}} \)
If using Combined Gas Law (with \( T_1 = 350 \, \text{K} \), \( T_2 = 400 \, \text{K} \)): \( \boldsymbol{\approx 4.57 \, \text{atm}} \) (depending on the correct interpretation of the typos in the problem)