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Worked Examples · Example 12.8

Q.A cylinder of fixed capacity 44.844.8 litres contains helium gas at standard temperature and pressure. What is the amount of heat needed to raise the temperature of the gas in the cylinder by 15.0 ∘C15.0\ ^\circ\text{C}? (R=8.31 J mol−1K−1R = 8.31\ \text{J mol}^{-1}\text{K}^{-1}).

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For a monatomic gas like helium in a fixed-volume container, the heat required is Q=nCVΔTQ = n C_V \Delta T. Using the ideal gas law at STP, n=2.00n = 2.00 mol, CV=32RC_V = \frac{3}{2}R, and ΔT=15.0 K\Delta T = 15.0\ \text{K}, the heat needed is Q=374 JQ = 374\ \text{J}.

The key here is to recognise that the cylinder has a fixed capacity — the volume does not change. That means the gas is heated at constant volume, not constant pressure. Many students instinctively reach for CpC_p (the molar heat capacity at constant pressure), but that would be wrong here. For a rigid container, the heat added goes entirely into increasing the internal energy, with no work done by the gas.

Helium is a monatomic ideal gas. For such a gas, the molar heat capacity at constant volume is CV=32RC_V = \frac{3}{2}R. This comes from the equipartition theorem: each translational degree of freedom contributes 12R\frac{1}{2}R to CVC_V, and a monatomic gas has three translational degrees of freedom.

Let’s work through the problem step by step.

  1. Find the number of moles of helium. At standard temperature and pressure (STP), T=273 KT = 273\ \text{K} and P=1.00 atm=1.013×105 PaP = 1.00\ \text{atm} = 1.013 \times 10^5\ \text{Pa}. The cylinder volume is V=44.8 litres=44.8×10−3 m3V = 44.8\ \text{litres} = 44.8 \times 10^{-3}\ \text{m}^3. Using the ideal gas law:

PV=nRTPV = nRT

n=PVRT=(1.013×105)(44.8×10−3)(8.31)(273)n = \frac{PV}{RT} = \frac{(1.013 \times 10^5)(44.8 \times 10^{-3})}{(8.31)(273)}

Compute stepwise:

PV=1.013×105×0.0448=4538 JPV = 1.013 \times 10^5 \times 0.0448 = 4538\ \text{J} (since Pa·m³ = J).

RT=8.31×273=2268 J/molRT = 8.31 \times 273 = 2268\ \text{J/mol}.

So n=45382268≈2.00 moln = \frac{4538}{2268} \approx 2.00\ \text{mol}.

Tip

At STP, 1 mole of any ideal gas occupies 22.4 litres. Here the volume is 44.8 litres, so directly n=44.8/22.4=2.00n = 44.8 / 22.4 = 2.00 mol — a quick check that saves calculation time.

  1. Determine the correct heat capacity. Since the volume is fixed, the heat required is Q=nCVΔTQ = n C_V \Delta T. For helium (monatomic), CV=32RC_V = \frac{3}{2}R. …

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