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Sagot :
Sure! Let's solve the problem step by step using the Ideal Gas Law, [tex]\( PV = nRT \)[/tex], where:
- [tex]\( P \)[/tex] is the pressure of the gas.
- [tex]\( V \)[/tex] is the volume of the gas.
- [tex]\( n \)[/tex] is the amount of gas (in moles).
- [tex]\( R \)[/tex] is the ideal gas constant.
- [tex]\( T \)[/tex] is the temperature of the gas in Kelvin.
Given data:
- [tex]\( n = 0.80 \, \text{mol} \)[/tex]
- [tex]\( V = 275 \, \text{mL} \)[/tex] which needs to be converted to [tex]\( \text{L} \)[/tex]
- [tex]\( P = 175 \, \text{kPa} \)[/tex]
- [tex]\( R = 8.814 \, \frac{ \text{L} \cdot \text{kPa} }{ \text{mol} \cdot \text{K} } \)[/tex]
First, convert the volume from milliliters (mL) to liters (L):
[tex]\[ 275 \, \text{mL} = 275 \times 10^{-3} \, \text{L} = 0.275 \, \text{L} \][/tex]
Now substitute all known values into the Ideal Gas Law equation, [tex]\( PV = nRT \)[/tex], and solve for [tex]\( T \)[/tex]:
[tex]\[ T = \frac{PV}{nR} \][/tex]
Substitute in the values:
[tex]\[ T = \frac{(175 \, \text{kPa}) (0.275 \, \text{L})}{(0.80 \, \text{mol}) (8.814 \, \frac{\text{L} \cdot \text{kPa}}{\text{mol} \cdot \text{K}})} \][/tex]
After performing all operations, you get the result:
[tex]\[ T \approx 6.825 \, \text{K} \][/tex]
Given the provided options:
- [tex]\( 4.6 \, \text{K} \)[/tex]
- [tex]\( 7.2 \, \text{K} \)[/tex]
- [tex]\( 61 \, \text{K} \)[/tex]
- [tex]\( 96 \, \text{K} \)[/tex]
The closest value to our calculated temperature of [tex]\( 6.825 \, \text{K} \)[/tex] is not perfectly matching any of these options exactly but appears closest to:
[tex]\[ 7.2 \, \text{K} \][/tex]
So, the temperature of the gas is approximately [tex]\( 7.2 \, \text{K} \)[/tex].
- [tex]\( P \)[/tex] is the pressure of the gas.
- [tex]\( V \)[/tex] is the volume of the gas.
- [tex]\( n \)[/tex] is the amount of gas (in moles).
- [tex]\( R \)[/tex] is the ideal gas constant.
- [tex]\( T \)[/tex] is the temperature of the gas in Kelvin.
Given data:
- [tex]\( n = 0.80 \, \text{mol} \)[/tex]
- [tex]\( V = 275 \, \text{mL} \)[/tex] which needs to be converted to [tex]\( \text{L} \)[/tex]
- [tex]\( P = 175 \, \text{kPa} \)[/tex]
- [tex]\( R = 8.814 \, \frac{ \text{L} \cdot \text{kPa} }{ \text{mol} \cdot \text{K} } \)[/tex]
First, convert the volume from milliliters (mL) to liters (L):
[tex]\[ 275 \, \text{mL} = 275 \times 10^{-3} \, \text{L} = 0.275 \, \text{L} \][/tex]
Now substitute all known values into the Ideal Gas Law equation, [tex]\( PV = nRT \)[/tex], and solve for [tex]\( T \)[/tex]:
[tex]\[ T = \frac{PV}{nR} \][/tex]
Substitute in the values:
[tex]\[ T = \frac{(175 \, \text{kPa}) (0.275 \, \text{L})}{(0.80 \, \text{mol}) (8.814 \, \frac{\text{L} \cdot \text{kPa}}{\text{mol} \cdot \text{K}})} \][/tex]
After performing all operations, you get the result:
[tex]\[ T \approx 6.825 \, \text{K} \][/tex]
Given the provided options:
- [tex]\( 4.6 \, \text{K} \)[/tex]
- [tex]\( 7.2 \, \text{K} \)[/tex]
- [tex]\( 61 \, \text{K} \)[/tex]
- [tex]\( 96 \, \text{K} \)[/tex]
The closest value to our calculated temperature of [tex]\( 6.825 \, \text{K} \)[/tex] is not perfectly matching any of these options exactly but appears closest to:
[tex]\[ 7.2 \, \text{K} \][/tex]
So, the temperature of the gas is approximately [tex]\( 7.2 \, \text{K} \)[/tex].
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