# A gas at 155 kPa and 25'C has an initial volume of 1.00 L. The pressure of the gas increases to 605 kPa as the temperature is raised to 125°C. What is the new volume?

Here, pressure prevails over temperature.

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To find the new volume of the gas, we can use the combined gas law, which states:

( P_1 \times V_1 / T_1 = P_2 \times V_2 / T_2 )

Given: ( P_1 = 155 ) kPa ( V_1 = 1.00 ) L ( T_1 = 25^\circ )C + 273.15 = 298.15 K ( P_2 = 605 ) kPa ( T_2 = 125^\circ )C + 273.15 = 398.15 K

We can rearrange the equation to solve for ( V_2 ):

( V_2 = (P_1 \times V_1 \times T_2) / (P_2 \times T_1) )

Substitute the values:

( V_2 = (155 \times 1.00 \times 398.15) / (605 \times 298.15) )

( V_2 = (61651.7) / (180520.75) )

( V_2 ≈ 0.34 ) L

Therefore, the new volume of the gas is approximately 0.34 L.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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