Physics Thermodynamics First Law of Thermodynamics (Delta Q = Delta U + Delta W) Single Correct MCQ
Published on: September 13, 2026

The specific heat of hydrogen gas at constant pressure is $c_{P} = 3.4 \times 10^{3} \text{cal/kg}^\circ \text{C}$ and at constant volume is $c_V = 2.4 \times 10^3 \text{ cal/kg}^\circ \text{C}.$ If one kilogram hydrogen gas is heated from $10^\circ \mathrm{C}$ to 20°C at constant pressure, the external work done on the gas to maintain it at constant pressure is

A
$10^{5}$ cal
B
$10^{4}$ cal
C
$10^{3}$ cal
D
$5 \times 10^{3}$ cal

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Text Solution

Verified by Experts
The correct answer is:
B

From FLOT Δ Δ Q = Δ Δ U + Δ Δ W

Work done at constant pressure $(\Delta W)_{P} = (\Delta Q)_{P} - \Delta U$

$(\Delta Q)_{P} - (\Delta Q)_{V}$ (As we know $(\Delta Q)_V = \Delta U$ )

Also $(\Delta Q)_{p} = m c_{p} \Delta T$ and $(\Delta Q)_V = m c_V \Delta T$

⇒ ⇒ $(\Delta W)_{p} = m(c_{p} - c_{V}) \Delta T$

⇒ ⇒ $(\Delta W)_{\mathrm{p}} = 1 \times (3.4 \times 10^{3} - 2.4 \times 10^{3}) \times 10 = 10^{4} \mathrm{cal}$

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