QuestionAn ideal gas has a volume of $1.50 \times 10^{-4} \mathrm{~m}^3$ at $22^{\circ} \mathrm{C}$ and $1.20 \times 10^5$ Pa. Find the temperature in Kelvin when pressure is $2.00 \times 10^5$ Pa.

Studdy Solution

STEP 1

Assumptions1. The initial volume of the gas is $1.50 \times10^{-4} m^3$ . The initial temperature of the gas is $22^{\circ}C$
3. The initial pressure of the gas is $1.20 \times10^{5} Pa$
4. The final pressure of the gas is $.00 \times10^{5} Pa$
5. The volume of the gas remains constant6. The gas behaves ideally

STEP 2

We will use the ideal gas law, which states that the pressure of a gas times its volume is proportional to the temperature of the gas.
$1V1/T1 =2V2/T2$

STEP 3

Since the volume of the gas remains constant, we can simplify the equation to $1/T1 =2/T2$

STEP 4

We need to solve for $2$ , the final temperature. Rearrange the equation to isolate $2$ $2 =2 \cdot1 /1$

STEP 5

Before we plug in the values, we need to convert the initial temperature from Celsius to Kelvin. The conversion formula is $(K) =(°C) +273.15$

STEP 6

Convert the initial temperature to Kelvin $1 =22°C +273.15 =295.15 K$

STEP 7

Now we can plug in the values for $2$ , $1$ , and $1$ into the equation $2 =2.00 \times10^{5} Pa \cdot295.15 K /1.20 \times10^{5} Pa$

STEP 8

Perform the calculation to find the final temperature $2 =2.00 \times10^{5} Pa \cdot295.15 K /1.20 \times10^{5} Pa =491.92 K$ The temperature at which the pressure of the gas is $2.00 \times10^{5} Pa$ is approximately $491.92 K$ .

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QuestionAn ideal gas has a volume of $1.50 \times 10^{-4} \mathrm{~m}^3$ at $22^{\circ} \mathrm{C}$ and $1.20 \times 10^5$ Pa. Find the temperature in Kelvin when pressure is $2.00 \times 10^5$ Pa.

Studdy Solution

STEP 1

STEP 2

We will use the ideal gas law, which states that the pressure of a gas times its volume is proportional to the temperature of the gas.
$1V1/T1 =2V2/T2$

STEP 3

Since the volume of the gas remains constant, we can simplify the equation to $1/T1 =2/T2$

STEP 4

We need to solve for $2$ , the final temperature. Rearrange the equation to isolate $2$ $2 =2 \cdot1 /1$

STEP 5

Before we plug in the values, we need to convert the initial temperature from Celsius to Kelvin. The conversion formula is $(K) =(°C) +273.15$

STEP 6

Convert the initial temperature to Kelvin $1 =22°C +273.15 =295.15 K$

STEP 7

Now we can plug in the values for $2$ , $1$ , and $1$ into the equation $2 =2.00 \times10^{5} Pa \cdot295.15 K /1.20 \times10^{5} Pa$

STEP 8

Perform the calculation to find the final temperature $2 =2.00 \times10^{5} Pa \cdot295.15 K /1.20 \times10^{5} Pa =491.92 K$ The temperature at which the pressure of the gas is $2.00 \times10^{5} Pa$ is approximately $491.92 K$ .

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QuestionAn ideal gas has a volume of 1.50×10−4 m31.50 \times 10^{-4} \mathrm{~m}^31.50×10−4 m3 at 22∘C22^{\circ} \mathrm{C}22∘C and 1.20×1051.20 \times 10^51.20×105 Pa. Find the temperature in Kelvin when pressure is 2.00×1052.00 \times 10^52.00×105 Pa.

Studdy Solution

STEP 1

STEP 2

We will use the ideal gas law, which states that the pressure of a gas times its volume is proportional to the temperature of the gas.1V1/T1=2V2/T21V1/T1 =2V2/T21V1/T1=2V2/T2

STEP 3

Since the volume of the gas remains constant, we can simplify the equation to1/T1=2/T21/T1 =2/T21/T1=2/T2

STEP 4

We need to solve for 222, the final temperature. Rearrange the equation to isolate 2222=2⋅1/12 =2 \cdot1 /12=2⋅1/1

STEP 5

Before we plug in the values, we need to convert the initial temperature from Celsius to Kelvin. The conversion formula is(K)=(°C)+273.15(K) =(°C) +273.15(K)=(°C)+273.15

STEP 6

Convert the initial temperature to Kelvin1=22°C+273.15=295.15K1 =22°C +273.15 =295.15 K1=22°C+273.15=295.15K

STEP 7

Now we can plug in the values for 222, 111, and 111 into the equation2=2.00×105Pa⋅295.15K/1.20×105Pa2 =2.00 \times10^{5} Pa \cdot295.15 K /1.20 \times10^{5} Pa2=2.00×105Pa⋅295.15K/1.20×105Pa

STEP 8

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QuestionAn ideal gas has a volume of $1.50 \times 10^{-4} \mathrm{~m}^3$ at $22^{\circ} \mathrm{C}$ and $1.20 \times 10^5$ Pa. Find the temperature in Kelvin when pressure is $2.00 \times 10^5$ Pa.

We will use the ideal gas law, which states that the pressure of a gas times its volume is proportional to the temperature of the gas.
$1V1/T1 =2V2/T2$

Since the volume of the gas remains constant, we can simplify the equation to $1/T1 =2/T2$

We need to solve for $2$ , the final temperature. Rearrange the equation to isolate $2$ $2 =2 \cdot1 /1$

Before we plug in the values, we need to convert the initial temperature from Celsius to Kelvin. The conversion formula is $(K) =(°C) +273.15$

Convert the initial temperature to Kelvin $1 =22°C +273.15 =295.15 K$

Now we can plug in the values for $2$ , $1$ , and $1$ into the equation $2 =2.00 \times10^{5} Pa \cdot295.15 K /1.20 \times10^{5} Pa$