Question

Determine the pressure change when a constant volume of gas at 2.50 atm is heated from 30.0 °C to 40.0 °C.

Answer 1

The pressure changes from 2.50 atm to 2.58 atm (an increase of approximately 0.08 atm) when the gas is heated from 30.0°C to 40.0°C.

Step-by-step explanation:

As the given volume of gas is constant, we can use Gay-Lussac's law to solve this problem as it relates pressure to temperature.

Gay-Lussac's law

`\boxed{\sf (P_1)/(T_1)=(P_2)/(T_2)}`

where:

• P₁ = Initial pressure
• T₁ = Initial temperature (in kelvins)
• P₂ = Final pressure
• T₂ = Final temperature (in kelvins)

First, we need to convert the given temperatures from Celsius to Kelvin by adding 273.15:

`\implies \sf T_1=30+273.15=303.15\;K`

`\implies \sf T_2=40+273.15=313.15\;K`

Therefore, the values to substitute into the equation are:

• P₁ = 2.50 atm
• T₁ = 303.15 K
• T₂ = 313.15 K

As we are solving for the final pressure, rearrange the equation to isolate P₂:

`\sf P_2=(P_1T_2)/(T_1)`

Substitute the values into the equation and solve for P₂:

`\implies \sf P_2=(2.50 \cdot 313.15)/(303.15)`

`\implies \sf P_2=(782.875)/(303.15)`

`\implies \sf P_2=2.58246742...`

`\implies \sf P_2=2.58\;atm\;(3\;s.f.)`

Therefore, the pressure changes from 2.50 atm to 2.58 atm (an increase of approximately 0.08 atm) when the gas is heated from 30.0°C to 40.0°C.