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From: Christopher Howard <christopher@librehacker.com>
Newsgroups: sci.physics
Subject: Re: ideal gas law - pressure and mass question
Date: Sat, 18 Apr 2026 08:45:27 -0800
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Thomas 'PointedEars' Lahn <PointedEars@web.de> writes:

> True.  With constant volume, the temperature will increase according to
>
>   ∆Q = m c_V ∆T  ⇔  ∆T = ∆Q/(m c_V),
>
> where Q is heat; m is the mass of the gas, and c_V is its specific heat
> capacity at constant volume.
>
> And since the equation of state (EOS) of an ideal gas, also known as "Ideal
> Gas Law", can be written
>
>   p V = N k_B T,
>
> if the volume V and number of particles N are constant, and the absolute
> temperature T increases, then the pressure p will increase proportionally to
> the change in T, therefore to the change in Q:
>
>   ∆p = (N k_B/V) ∆T ∝ ∆T ∝ ∆Q.
>

Intuitively, it makes sense to me that if I have a greater mass of gas
involved, therefore the pressure will increase more slowly with the same
amount of applied heat.

I can see certainly that temperature will increase more slowly, because,
as you point out, ∆T = ∆Q/(m c_V).

However, in the equation p V = N k_B T, the number of particles, N, does
change with an increase in mass, correct? I think if we assumed we were
dealing with air, with a molar mass of about 30 g/mol, or 3x10⁻² kg/mol,
then the equation converts to this, correct?:

M = N × 3x10⁻² kg/mol
N = M / (3x10⁻² kg/mol)

Δp = (M / (3x10⁻² kg/mol)) × (R / V) × ΔT
Δp = (M / (3x10⁻² kg/mol)) × (R / V) × (∆Q / M c_air)

And therefore the mass terms (M) cancel out, correct...?

-- 
Christopher Howard