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Determine the pressure change when a constant volume of gas at 1.00 atm is heated from 20.0C to 40C

a
2.00 C
b
1.06 atm
c
1.06 K
d
2.00 atm
e
.936 K
f
0.50 atm
g
.936 atm
h
0.50C

1 Answer

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The pressure change when a constant volume of gas is heated from 20.0°C to 40.0°C is an increase of 0.07 atm.

To determine the pressure change when a constant volume of gas at 1.00 atm is heated from 20.0°C to 40.0°C, we can use the ideal gas law equation: PV = nRT.

Since the volume is constant, we can simplify the equation to P1/T1 = P2/T2, where P1 and T1 are the initial pressure and temperature, and P2 and T2 are the final pressure and temperature.

Using the given temperatures in Kelvin, T1 = 20.0 + 273 = 293 K and T2 = 40.0 + 273 = 313 K.

Substituting these values into the equation, we get P1/293 = P2/313. Solving for P2, we have P2 = P1 * (T2/T1).

Plugging in the values, P2 = 1.00 atm * (313/293) = 1.07 atm.

Therefore, the pressure change when the gas is heated from 20.0°C to 40.0°C is an increase of 0.07 atm.

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