Q.Molar volume is the volume occupied by $1\ \text{mol}$ of any (ideal) gas at standard temperature and pressure (STP: $1$ atmospheric pressure, $0\ ^\circ\text{C}$). Show that it is $22.4$ litres.

Showing the 12 most recent of 27 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.Which of the following is the value of the universal gas constant (R)?

   (a) 8.13 J mol^-1 K^-1

   (b) 8.31 J mol^-1 K^-1

   (c) 8.31 J^-1 mol^-1 K^-1

   (d) 8.13 J^-1 mol^-1 K^-1

   ›Reveal solutionSolution

   The universal gas constant R = 8.31 J mol⁻¹ K⁻¹ (commonly rounded from 8.314 J mol⁻¹ K⁻¹).

   From the ideal gas equation PV = nRT, R is the constant of proportionality connecting pressure, volume, amount of substance, and temperature, and is the same for all ideal gases (hence 'universal'). Its measured value is R = 8.314 J mol⁻¹ K⁻¹ ≈ 8.31 J mol⁻¹ K⁻¹ (equivalently ≈ 1.987 cal mol⁻¹ K⁻¹, or 0.0821 L·atm mol⁻¹ K⁻¹ in other unit systems). Options (c) and (d) are wrong because they invert the unit (J⁻¹ instead of J), and option (a) simply has the digits transposed (8.13 instead of 8.31).

   ✓Final answer

   The correct option is (b) 8.31 J mol⁻¹ K⁻¹.

2. CBSE 2026Set ANNUAL1 mark
   Q.Write the ideal gas equation.

   ›Reveal solutionSolution

   The ideal gas equation is PV = nRT.

   Combining Boyle's law (P ∝ 1/V at constant T), Charles's law (V ∝ T at constant P), and Avogadro's law (V ∝ n at constant P, T) gives the single equation of state for an ideal gas: PV = nRT, where P is the pressure, V the volume, n the number of moles of gas, R the universal gas constant (8.31 J mol⁻¹ K⁻¹), and T the absolute temperature (in kelvin). Equivalently, in terms of the total number of molecules N and Boltzmann's constant k_B (R = N_A k_B), it can be written PV = Nk_BT.

   ✓Final answer

   PV = nRT.

3. CBSE 2026Set ANNUAL1 markMCQ
   Q.According to ideal gas equation:

   (a) PV = μRT

   (b) PV^r = Constant

   (c) P^rV = RT

   (d) PV = μR/T

   ›Reveal solutionSolution

   The ideal gas equation of state is PV = mu*RT, connecting pressure, volume, the number of moles, the universal gas constant, and absolute temperature.

   The ideal gas equation is derived by combining Boyle's law (PV = constant at constant T), Charles's law (V/T = constant at constant P), and Avogadro's law (V proportional to number of moles at constant P and T) into a single equation of state:

   PV = muRT

   where:

   P = pressure of the gas

   V = volume of the gas

   mu = number of moles of gas

   R = universal gas constant (8.314 J/mol K)

   T = absolute temperature (in kelvin)

   The other options are not correct: PV^gamma = constant is the equation for an adiabatic process (not the general equation of state), and the other two forms mix up the powers/placement of the variables incorrectly.

   ✓Final answer

   The correct option is (a) PV = mu*RT — this is the standard ideal gas equation of state.

4. CBSE 2026Set ANNUAL1 markMCQ
   Q.If the internal energy of an ideal gas U and volume V are doubled, then the pressure:

   (a) halves

   (b) doubles

   (c) increases four times

   (d) remains same

   ›Reveal solutionSolution

   Since U is proportional to T for an ideal gas, doubling U doubles T; combined with V also doubling, the pressure P = mu*RT/V is unchanged because the factor of 2 in T is exactly cancelled by the factor of 2 in V.

   For an ideal gas (fixed number of moles mu), the internal energy is

   U = (f/2)muR*T

   where f is the number of degrees of freedom. This shows U is directly proportional to the absolute temperature T.

   Step 1: If U becomes 2U, then since U is proportional to T, the temperature must also become 2T (with mu and f unchanged).

   Step 2: The ideal gas equation is PV = muRT, so P = muRT/V.

   Step 3: With the new values T' = 2T and V' = 2V,

   P' = muR(2T)/(2V) = muRT/V = P (unchanged)

   So the doubling of temperature (from doubled U) is exactly cancelled by the doubling of volume, leaving the pressure the same as before.

   ✓Final answer

   The correct option is (d) remains same — U proportional to T means doubling U doubles T, and this doubled T combined with the doubled V leaves P = mu*RT/V unchanged.

5. CBSE 2025Set ANNUAL1 markMCQ
   Q.The dimension of universal gas constant R is the same as that of (A) energy (B) heat capacity (C) molar heat capacity (D) temperature

   ›Reveal solutionSolution

   The universal gas constant R has the same dimension as molar heat capacity.

   From the ideal gas law $PV = nRT$:

   $R = \dfrac{PV}{nT}$

   Dimensionally, $[PV] =$ energy $= [ML^2T^{-2}]$, so:

   $[R] = \dfrac{[ML^2T^{-2}]}{[\text{mol}][K]}$

   This is energy per mole per kelvin — exactly the definition of molar heat capacity (heat required to raise the temperature of one mole of substance by one kelvin), which also has units J mol⁻¹ K⁻¹. (Plain "heat capacity" has units J/K, without the mole term, so it does not match; "energy" alone is J, also not matching; "temperature" is obviously different.)

   ✓Final answer

   (C) molar heat capacity.

6. CBSE 2025Set ANNUAL1 markMCQ
   Q.The graph between Volume and Temperature in Charles' law is:

   (a) a straight line

   (b) an ellipse

   (c) a parabola

   (d) a circle

   ›Reveal solutionSolution

   Charles' Law (V proportional to T at constant pressure) gives a straight-line V-T graph passing through the origin, when temperature is measured on the absolute (Kelvin) scale.

   Charles' Law: for a fixed mass of gas held at constant pressure, V/T = constant, or equivalently V = (constant) x T.

   This is exactly the equation of a straight line y = mx through the origin, with V playing the role of y, T playing the role of x, and the constant playing the role of the slope m.

   So as absolute temperature increases, volume increases in direct (linear) proportion -- the graph is a straight line, and extrapolating it back would meet the T-axis at absolute zero (T=0 K), which is in fact how the Kelvin scale's zero point was historically deduced.

   ✓Final answer

   The correct option is (a) a straight line.

7. CBSE 2025Set hz1 markMCQ
   Q.Two vessels A and B of the same size are at the same temperature, one of them holds 1Kg of H2 gas and the other hold 1Kg of N2 gas. Which of the vessels contains more molecules?

   (a) A only

   (b) B only

   (c) Both

   (a) and

   (b)

   (d) None of them

   ›Reveal solutionSolution

   Number of molecules = (mass / molar mass) x Avogadro's number. Since H2 has a much smaller molar mass than N2, 1 kg of H2 contains many more molecules than 1 kg of N2, so vessel A (H2) has more molecules.

   The number of molecules in a sample is N = n x N_A, where n = (given mass)/(molar mass) is the number of moles and N_A = 6.022 x 10^23 /mol is Avogadro's number.

   For vessel A, holding 1 kg = 1000 g of H2 (molar mass = 2 g/mol):

   n(H2) = 1000/2 = 500 mol

   N(H2) = 500 x N_A

   For vessel B, holding 1 kg = 1000 g of N2 (molar mass = 28 g/mol):

   n(N2) = 1000/28 approx 35.7 mol

   N(N2) approx 35.7 x N_A

   Since 500 mol is far greater than 35.7 mol, vessel A (H2) contains many more molecules than vessel B (N2), even though both vessels hold the same mass (1 kg) at the same temperature and same volume (which only fixes their pressures to be different, via PV = nRT, not their molecule counts).

   ✓Final answer

   The correct option is (a) A only.

8. CBSE 2025Set ANNUAL1 markMCQ
   Q.For an ideal gas, $PV = XT$, where $X$ is a constant, $X$ must be proportional to

   (a) mass of the gas molecule

   (b) absolute temperature

   (c) number of gas molecules in the vessel

   (d) kinetic energy of the gas.

   ›Reveal solutionSolution

   The ideal gas equation is $PV=nRT$, where $n$ is the number of moles of gas and $R$ is the universal gas constant (a true constant, same for all gases).

   Comparing the given equation $PV=XT$ with $PV=nRT$ term by term:

   $X = nR$

   Since $R$ is a universal constant, $X$ is directly proportional to $n$, the number of moles of gas present, which in turn is directly proportional to the number of gas molecules in the vessel (via Avogadro's number). $X$ does not depend on the mass of a single molecule or on the kinetic energy of the gas.

   ✓Final answer

   (c) number of gas molecules in the vessel

9. CBSE 2025Set ANNUAL1 markMCQ
   Q.At constant temperature, pressure of a given mass of a gas varies inversely with its volume. This statement is known as

   (a) (A) Boyle's law

   (b) (B) Charles' law

   (c) (C) Kelvin's law

   (d) (D) Gay Lussac's law

   ›Reveal solutionSolution

   [!TLDR]

   (A) Boyle's law

   Why

   This is the direct statement of Boyle's law (PV = constant at constant T).

   [!ANSWER]

   (A) Boyle's law

10. CBSE 2024Set ANNUAL1 markMCQ
    Q.Volume of a given mass of gas at constant pressure is (A) inversely proportional to absolute temperature (B) proportional to absolute temperature (C) proportional to temperature (D) inversely proportional to temperature

    ›Reveal solutionSolution

    By Charles's law, $V\propto T$ (absolute temperature) at constant pressure.

    For an ideal gas, $PV=nRT$. At constant pressure and fixed amount of gas ($n$), $V=\dfrac{nR}{P}T$, i.e. $V\propto T$ where $T$ must be the absolute (Kelvin) temperature — this direct proportionality only holds true using the Kelvin scale, not the Celsius scale.

    ✓Final answer

    (B) proportional to absolute temperature.

11. CBSE 2024Set SET-AP55001 mark
    Q.What is Boyle's law?

    ›Reveal solutionSolution

    Boyle's law states that for a fixed mass of gas at constant temperature, pressure is inversely proportional to volume: PV = constant.

    Formally, for a fixed quantity (mass/moles) of an ideal gas held at constant temperature T:

    P ∝ 1/V ⟹ PV = constant

    Equivalently, if the gas changes from state (P1, V1) to (P2, V2) at the same temperature, then P1V1 = P2V2. Physically, compressing the gas into a smaller volume (at the same temperature, so same average molecular speed) forces the same number of molecules to collide with the container walls more frequently, raising the pressure proportionally.

    ✓Final answer

    Boyle's law: PV = constant at constant temperature (for a fixed mass of gas).

12. CBSE 2024Set SET-NDP60001 markMCQ
    Q.What is the relationship between pressure P1 & P2 in the given V-T diagram?

    (a) P1 > P2

    (b) P2 > P1

    (c) P1 = P2

    (d) Cannot be determined

    ›Reveal solutionSolution

    On an isochoric-family V–T plot, a steeper line means a lower pressure, so the steeper line (P2, closer to the V-axis) has the smaller pressure, making $P_1 > P_2$.

    For $n$ moles of an ideal gas, $PV = nRT$, so at fixed pressure $P$ and fixed amount $n$,

    $V = \left(\dfrac{nR}{P}\right) T$

    This is the equation of a straight line through the origin on a V–T graph, with slope $= nR/P$. Since $n$ and $R$ are the same for both lines, the slope is inversely proportional to the pressure: a larger slope (a line closer to the V-axis, i.e. rising more steeply) corresponds to a smaller pressure, and a smaller slope (a line closer to the T-axis, i.e. flatter) corresponds to a larger pressure.

    In the given figure, line P2 is the steeper one (closer to the V-axis), so it has the larger slope and therefore the smaller pressure. Line P1 is the flatter one (closer to the T-axis), so it has the smaller slope and the larger pressure. Hence $P_1 > P_2$.

    ✓Final answer

    The correct option is (a) $P_1 > P_2$ — since line P2 is steeper (larger slope), it corresponds to the smaller pressure, so $P_1$ must be greater than $P_2$.