Question

1. a 1.2l sample of gas at 0.95 atm and 280k is changed to 1.4 atm and 310k what is the new volume ?

Explanation:

Step1: Identify the gas law

This problem involves gas pressure and volume change, so we use Boyle's Law, which states that for a fixed amount of gas at constant temperature, \( P_1V_1 = P_2V_2 \). Here, we assume temperature is constant (not mentioned to change), initial pressure \( P_1 = 0.95 \, \text{atm} \), initial volume \( V_1 = 1.2 \, \text{L} \), final pressure \( P_2 = 1.4 \, \text{atm} \), and we need to find final volume \( V_2 \).

Step2: Rearrange Boyle's Law formula

From \( P_1V_1 = P_2V_2 \), we solve for \( V_2 \): \( V_2=\frac{P_1V_1}{P_2} \)

Step3: Substitute the values

Substitute \( P_1 = 0.95 \, \text{atm} \), \( V_1 = 1.2 \, \text{L} \), \( P_2 = 1.4 \, \text{atm} \) into the formula:
\( V_2=\frac{0.95 \times 1.2}{1.4} \)

Step4: Calculate the value

First, calculate the numerator: \( 0.95\times1.2 = 1.14 \)
Then divide by the denominator: \( V_2=\frac{1.14}{1.4}\approx0.814 \, \text{L} \) (Wait, there was a typo in the problem? Wait, the initial volume was maybe 1.2 L, initial pressure 0.95 atm, final pressure 1.4 atm, but also there's a 280K and 310K? Wait, maybe I misread. Wait, the problem says: "A 1.2 L sample of gas at 0.95 atm and 280K is changed to 1.4 atm and 310K what is the new volume?" Oh! I missed the temperature change. So it's combined gas law: \( \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} \)

Let's correct:

Step1: Identify the gas law (combined gas law)

Since both pressure and temperature change, use combined gas law: \( \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} \), where \( P_1 = 0.95 \, \text{atm} \), \( V_1 = 1.2 \, \text{L} \), \( T_1 = 280 \, \text{K} \), \( P_2 = 1.4 \, \text{atm} \), \( T_2 = 310 \, \text{K} \), find \( V_2 \).

Step2: Rearrange the formula for \( V_2 \)

\( V_2=\frac{P_1V_1T_2}{P_2T_1} \)

Step3: Substitute the values

\( V_2=\frac{0.95 \times 1.2 \times 310}{1.4 \times 280} \)

Step4: Calculate numerator and denominator

Numerator: \( 0.95\times1.2 = 1.14 \); \( 1.14\times310 = 353.4 \)
Denominator: \( 1.4\times280 = 392 \)
Then \( V_2=\frac{353.4}{392}\approx0.901 \, \text{L} \) (Wait, maybe the initial volume was 1.2 L, initial pressure 0.95 atm, initial temp 280K; final pressure 1.4 atm, final temp 310K. Let's do the calculation properly.

Wait, let's re-express the problem correctly (from the image, it's: "11) A 1.2 L sample of gas at 0.95 atm and 280K is changed to 1.4 atm and 310K what is the new volume?")

So combined gas law: \( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \)

Solving for \( V_2 \):

\( V_2 = \frac{P_1V_1T_2}{P_2T_1} \)

Plug in the numbers:

\( P_1 = 0.95 \, \text{atm} \), \( V_1 = 1.2 \, \text{L} \), \( T_1 = 280 \, \text{K} \)

\( P_2 = 1.4 \, \text{atm} \), \( T_2 = 310 \, \text{K} \)

\( V_2 = \frac{0.95 \times 1.2 \times 310}{1.4 \times 280} \)

Calculate numerator: \( 0.95 \times 1.2 = 1.14 \); \( 1.14 \times 310 = 353.4 \)

Denominator: \( 1.4 \times 280 = 392 \)

\( V_2 = \frac{353.4}{392} \approx 0.901 \, \text{L} \) (approximately 0.90 L)

Wait, but maybe the initial volume was 1.2 L, initial pressure 0.95 atm, initial temp 280K; final pressure 1.4 atm, final temp 310K. So the calculation is as above.

Answer:

The new volume is approximately \(\boxed{0.90 \, \text{L}}\) (or more precisely \(\approx 0.901 \, \text{L}\))