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Vladimir79 [104]

4 years ago

12

At a temperature of 393 K, the pressure of a sample of nitrogen is 1.07 atm. What will the pressure be at a temperature of 478 K

? (Assume constant volume)

1 answer:

Rainbow [258]4 years ago

6 0

Answer:

1.30atm

Explanation:

P1/T1 = P2/T2

1.07/393 = P2/478

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Montano1993 [528]

Answer:

$\boxed {\boxed {\sf 232.9 \textdegree C}}$

Explanation:

This question asks us to find the temperature change given a volume change. We will use Charles's Law, which states the volume of a gas is directly proportional to the temperature. The formula is:

$\frac {V_1}{T_1}= \frac{V_2}{T_2}$

The volume of the gas starts at 250 milliliters and the temperature is 137 °C.

$\frac{250 \ mL}{137 \textdegree C}= \frac{V_2}{T_2}$

The volume of the gas is increased to 425 milliliters, but the temperature is unknown.

$\frac{250 \ mL}{137 \textdegree C}= \frac{425 \ mL}{T_2}$

We are solving for the new temperature, so we must isolate the variable T₂. First, cross multiply. Multiply the first numerator and second denominator, then multiply the first denominator and second numerator.

$250 \ mL * T_2 = 137 \textdegree C * 425 \ mL$

Now the variable is being multiplied by 250 milliliters. The inverse of multiplication is division. Divide both sides of the equation by 250 mL.

$\frac{250 \ mL * T_2}{250 \ mL}=\frac{ 137 \textdegree C * 425 \ mL}{250 \ mL}$

$T_2=\frac{ 137 \textdegree C * 425 \ mL}{250 \ mL}$

The units of milliliters (mL) cancel.

$T_2=\frac{ 137 \textdegree C * 425 }{250 }$

$T_2= \frac{58225}{250} \textdegree C$

$T_2=232.9 \textdegree C$

The temperature changes to <u>232.9 degrees Celsius.</u>

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